Comprehensive Notes on Ordinary Differential Equations

Theoretical Foundations of Ordinary Differential Equations

  • General Structure of Linear Ordinary Differential Equations (ODEs):

    • An nn-th order linear ODE with constant coefficients is represented as:         a0dnydxn+a1dn−1ydxn−1+⋯+an−1dydx+any=X(x)a_0 \frac{d^n y}{dx^n} + a_1 \frac{d^{n-1} y}{dx^{n-1}} + \dots + a_{n-1} \frac{dy}{dx} + a_n y = X(x)
    • Using the differential operator D=ddxD = \frac{d}{dx}, the equation is written in operator form as:         f(D)y=X(x)f(D)y = X(x)         where f(D)=a0Dn+a1Dn−1+⋯+an−1D+anf(D) = a_0 D^n + a_1 D^{n-1} + \dots + a_{n-1} D + a_n.

  • Composition of the Complete Solution:

    • The complete general solution y(x)y(x) to a non-homogeneous linear ODE consists of two distinct parts:         y(x)=yc(x)+yp(x)y(x) = y_c(x) + y_p(x)
    • Complementary Function (yc(x)y_c(x)): The solution to the homogeneous linear equation f(D)y=0f(D)y = 0.
    • Particular Integral (yp(x)y_p(x)): Any specific solution that satisfies the non-homogeneous equation f(D)y=X(x)f(D)y = X(x).

  • Determination of the Complementary Function (yc(x)y_c(x)):

    • The auxiliary equation (A.E.) is formed by substituting D=mD = m into the operator polynomial: f(m)=0f(m) = 0.
    • Case 1: Real and Distinct Roots (m1,m2,…,mnm_1, m_2, \dots, m_n):yc=C1em1x+C2em2x+⋯+Cnemnxy_c = C_1 e^{m_1 x} + C_2 e^{m_2 x} + \dots + C_n e^{m_n x}
    • Case 2: Real and Equal Roots (m1=m2=mm_1 = m_2 = m):yc=(C1+C2x)emxy_c = (C_1 + C_2 x) e^{m x}
    • Case 3: Complex Conjugate Roots (m=α±iβm = \alpha \pm i\beta):yc=eαx(C1cos⁡(βx)+C2sin⁡(βx))y_c = e^{\alpha x} (C_1 \cos(\beta x) + C_2 \sin(\beta x))
    • Case 4: Repeated Complex Conjugate Roots (m=α±iβm = \alpha \pm i\beta repeated twice):yc=eαx((C1+C2x)cos⁡(βx)+(C3+C4x)sin⁡(βx))y_c = e^{\alpha x} ((C_1 + C_2 x)\cos(\beta x) + (C_3 + C_4 x)\sin(\beta x))

Standard Evaluation Rules for Particular Integrals

  • Type I: Exponential Forcing Functions (X(x)=eaxX(x) = e^{ax}):

    • yp=1f(D)eax=1f(a)eaxy_p = \frac{1}{f(D)} e^{ax} = \frac{1}{f(a)} e^{ax}, provided $f(a) eq 0$.
    • If f(a)=0f(a) = 0 (case of failure), differentiate denominator with respect to DD and multiply numerator by xx:         yp=x1f′(a)eaxy_p = x \frac{1}{f'(a)} e^{ax}
    • If f′(a)=0f'(a) = 0, repeat to obtain yp=x21f′′(a)eaxy_p = x^2 \frac{1}{f''(a)} e^{ax}.

  • Type II: Sinusoidal Forcing Functions (X(x)=sin⁡(ax)X(x) = \sin(ax) or X(x)=cos⁡(ax)X(x) = \cos(ax)):

    • Substitute D2=−a2D^2 = -a^2 into f(D)f(D):         yp=1f(D2)sin⁡(ax)=1f(−a2)sin⁡(ax)y_p = \frac{1}{f(D^2)} \sin(ax) = \frac{1}{f(-a^2)} \sin(ax), provided $f(-a^2) eq 0$.
    • If f(−a2)=0f(-a^2) = 0, multiply by xx and differentiate operator denominator:         yp=x1f′(D)sin⁡(ax)y_p = x \frac{1}{f'(D)} \sin(ax)

  • Type III: Polynomial Forcing Functions (X(x)=xmX(x) = x^m):

    • Expand f(D)−1f(D)^{-1} in ascending powers of DD using binomial series expansions:         (1−D)−1=1+D+D2+D3+…(1 - D)^{-1} = 1 + D + D^2 + D^3 + \dots(1+D)−1=1−D+D2−D3+…(1 + D)^{-1} = 1 - D + D^2 - D^3 + \dots(1−D)−2=1+2D+3D2+4D3+…(1 - D)^{-2} = 1 + 2D + 3D^2 + 4D^3 + \dots(1+D)−2=1−2D+3D2−4D3+…(1 + D)^{-2} = 1 - 2D + 3D^2 - 4D^3 + \dots
    • Apply the expansion operator directly to xmx^m.

  • Type IV: Product Forcing Functions (X(x)=eaxV(x)X(x) = e^{ax} V(x)):

    • Shift the exponential factor out of the operator by replacing DD with D+aD + a:         yp=1f(D)[eaxV(x)]=eax1f(D+a)V(x)y_p = \frac{1}{f(D)} [e^{ax} V(x)] = e^{ax} \frac{1}{f(D + a)} V(x)

  • Trigonometric Identity Simplifications:

    • cos⁡2(x)=1+cos⁡(2x)2\cos^2(x) = \frac{1 + \cos(2x)}{2}
    • sin⁡2(x)=1−cos⁡(2x)2\sin^2(x) = \frac{1 - \cos(2x)}{2}

Special Classes of Higher-Order Differential Equations

  • Cauchy-Euler Linear Differential Equations:

    • Standard second-order equation form:         x2d2ydx2+a1xdydx+a2y=X(x)x^2 \frac{d^2y}{dx^2} + a_1 x \frac{dy}{dx} + a_2 y = X(x)
    • Transformation Substitution: Set x=ezx = e^z, which gives z=ln⁡(x)z = \ln(x).
    • Operator Reductions: Let θ=ddz\theta = \frac{d}{dz}.         xdydx=θyx \frac{dy}{dx} = \theta yx2d2ydx2=θ(θ−1)yx^2 \frac{d^2y}{dx^2} = \theta(\theta - 1) y

  • Legendre Linear Differential Equations:

    • Standard second-order equation form:         (ax+b)2d2ydx2+c1(ax+b)dydx+c2y=X(x)(ax + b)^2 \frac{d^2y}{dx^2} + c_1 (ax + b) \frac{dy}{dx} + c_2 y = X(x)
    • Transformation Substitution: Set ax+b=ezax + b = e^z, which gives z=ln⁡(ax+b)z = \ln(ax + b).
    • Operator Reductions: Let θ=ddz\theta = \frac{d}{dz}.         (ax+b)dydx=aθy(ax + b) \frac{dy}{dx} = a \theta y(ax+b)2d2ydx2=a2θ(θ−1)y(ax + b)^2 \frac{d^2y}{dx^2} = a^2 \theta(\theta - 1) y

  • Method of Variation of Parameters:

    • Used for general non-homogeneous linear second-order ODEs: d2ydx2+P(x)dydx+Q(x)y=R(x)\frac{d^2y}{dx^2} + P(x) \frac{dy}{dx} + Q(x) y = R(x).
    • Step 1: Obtain complementary solution yc=C1y1(x)+C2y2(x)y_c = C_1 y_1(x) + C_2 y_2(x).
    • Step 2: Compute Wronskian determinant:         W(y1,y2)=y1y2′−y1′y2=(y1y2y1′y2′)W(y_1, y_2) = y_1 y_2' - y_1' y_2 = \begin{pmatrix} y_1 & y_2 \\ y_1' & y_2' \end{pmatrix}
    • Step 3: Calculate variable parameters u(x)u(x) and v(x)v(x):         u(x)=−∫y2(x)R(x)W(y1,y2)dxu(x) = -\int \frac{y_2(x) R(x)}{W(y_1, y_2)} dxv(x)=∫y1(x)R(x)W(y1,y2)dxv(x) = \int \frac{y_1(x) R(x)}{W(y_1, y_2)} dx
    • Step 4: Form particular integral yp=u(x)y1(x)+v(x)y2(x)y_p = u(x) y_1(x) + v(x) y_2(x).
    • Step 5: Write general solution y=C1y1(x)+C2y2(x)+ypy = C_1 y_1(x) + C_2 y_2(x) + y_p.

Solutions to Part-A Differential Equations Problems

  • Problem A.1: Solve y′+2y=e−xy' + 2y = e^{-x}

    • This is a first-order linear ODE: dydx+2y=e−x\frac{dy}{dx} + 2y = e^{-x}.
    • Integrating Factor: I.F.=e∫2dx=e2xI.F. = e^{\int 2 dx} = e^{2x}.
    • General Solution:         y⋅e2x=∫e−x⋅e2xdx+Cy \cdot e^{2x} = \int e^{-x} \cdot e^{2x} dx + Cy⋅e2x=∫exdx+C=ex+Cy \cdot e^{2x} = \int e^x dx + C = e^x + Cy(x)=e−x+Ce−2xy(x) = e^{-x} + C e^{-2x}

  • Problem A.2: Solve (D2−2D−3)y=5(D^2 - 2D - 3)y = 5

    • Auxiliary Equation: m2−2m−3=0  ⟹  (m−3)(m+1)=0  ⟹  m=3,−1m^2 - 2m - 3 = 0 \implies (m - 3)(m + 1) = 0 \implies m = 3, -1.
    • Complementary Function: yc=C1e3x+C2e−xy_c = C_1 e^{3x} + C_2 e^{-x}.
    • Particular Integral:         yp=1D2−2D−3(5e0x)=502−2(0)−3=−53y_p = \frac{1}{D^2 - 2D - 3} (5e^{0x}) = \frac{5}{0^2 - 2(0) - 3} = -\frac{5}{3}
    • General Solution: y=C1e3x+C2e−x−53y = C_1 e^{3x} + C_2 e^{-x} - \frac{5}{3}.

  • Problem A.3: Solve (D2+6D+9)y=0(D^2 + 6D + 9)y = 0

    • Auxiliary Equation: m2+6m+9=0  ⟹  (m+3)2=0  ⟹  m=−3,−3m^2 + 6m + 9 = 0 \implies (m + 3)^2 = 0 \implies m = -3, -3.
    • General Solution: y=(C1+C2x)e−3xy = (C_1 + C_2 x) e^{-3x}.

  • Problem A.4: Solve (D2+4)y=sin⁡(2x)(D^2 + 4)y = \sin(2x)

    • Auxiliary Equation: m2+4=0  ⟹  m=±2im^2 + 4 = 0 \implies m = \pm 2i.
    • Complementary Function: yc=C1cos⁡(2x)+C2sin⁡(2x)y_c = C_1 \cos(2x) + C_2 \sin(2x).
    • Particular Integral: Substitute D2→−22=−4D^2 \to -2^2 = -4, leading to zero denominator.         yp=x12Dsin⁡(2x)=x2∫sin⁡(2x)dx=−xcos⁡(2x)4y_p = x \frac{1}{2D} \sin(2x) = \frac{x}{2} \int \sin(2x) dx = -\frac{x \cos(2x)}{4}
    • General Solution: y=C1cos⁡(2x)+C2sin⁡(2x)−xcos⁡(2x)4y = C_1 \cos(2x) + C_2 \sin(2x) - \frac{x \cos(2x)}{4}.

  • Problem A.5: Solve (D2+4)y=cos⁡2(x)(D^2 + 4)y = \cos^2(x)

    • Rewrite forcing term: cos⁡2(x)=12+12cos⁡(2x)\cos^2(x) = \frac{1}{2} + \frac{1}{2} \cos(2x).
    • Complementary Function: yc=C1cos⁡(2x)+C2sin⁡(2x)y_c = C_1 \cos(2x) + C_2 \sin(2x).
    • Particular Integrals:         yp1=1D2+4(12e0x)=18y_{p1} = \frac{1}{D^2 + 4} \left(\frac{1}{2} e^{0x}\right) = \frac{1}{8}yp2=1D2+4(12cos⁡(2x))=xsin⁡(2x)8y_{p2} = \frac{1}{D^2 + 4} \left(\frac{1}{2} \cos(2x)\right) = \frac{x \sin(2x)}{8}
    • General Solution: y=C1cos⁡(2x)+C2sin⁡(2x)+18+xsin⁡(2x)8y = C_1 \cos(2x) + C_2 \sin(2x) + \frac{1}{8} + \frac{x \sin(2x)}{8}.

  • Problem A.6: Solve (D2+4)y=x4(D^2 + 4)y = x^4

    • Complementary Function: yc=C1cos⁡(2x)+C2sin⁡(2x)y_c = C_1 \cos(2x) + C_2 \sin(2x).
    • Particular Integral:         yp=14(1+D24)−1x4=14(1−D24+D416)x4y_p = \frac{1}{4} \left(1 + \frac{D^2}{4}\right)^{-1} x^4 = \frac{1}{4} \left(1 - \frac{D^2}{4} + \frac{D^4}{16}\right) x^4yp=14(x4−12x24+2416)=x44−3x24+38y_p = \frac{1}{4} \left(x^4 - \frac{12x^2}{4} + \frac{24}{16}\right) = \frac{x^4}{4} - \frac{3x^2}{4} + \frac{3}{8}
    • General Solution: y=C1cos⁡(2x)+C2sin⁡(2x)+x44−3x24+38y = C_1 \cos(2x) + C_2 \sin(2x) + \frac{x^4}{4} - \frac{3x^2}{4} + \frac{3}{8}.

  • Problem A.7: Solve (D2+2D+1)y=e−x+3(D^2 + 2D + 1)y = e^{-x} + 3

    • Auxiliary Equation: m2+2m+1=0  ⟹  (m+1)2=0  ⟹  m=−1,−1m^2 + 2m + 1 = 0 \implies (m + 1)^2 = 0 \implies m = -1, -1.
    • Complementary Function: yc=(C1+C2x)e−xy_c = (C_1 + C_2 x) e^{-x}.
    • Particular Integrals:         yp1=1(D+1)2e−x=x212e−x=x2e−x2y_{p1} = \frac{1}{(D + 1)^2} e^{-x} = x^2 \frac{1}{2} e^{-x} = \frac{x^2 e^{-x}}{2}yp2=1(D+1)2(3e0x)=3y_{p2} = \frac{1}{(D + 1)^2} (3e^{0x}) = 3
    • General Solution: y=(C1+C2x)e−x+x2e−x2+3y = (C_1 + C_2 x) e^{-x} + \frac{x^2 e^{-x}}{2} + 3.

  • Problem A.8: Solve d2ydx2−7dydx+12y=0\frac{d^2y}{dx^2} - 7\frac{dy}{dx} + 12y = 0

    • Auxiliary Equation: m2−7m+12=0  ⟹  (m−3)(m−4)=0  ⟹  m=3,4m^2 - 7m + 12 = 0 \implies (m - 3)(m - 4) = 0 \implies m = 3, 4.
    • General Solution: y=C1e3x+C2e4xy = C_1 e^{3x} + C_2 e^{4x}.

Solutions to Part-B Differential Equations Problems

  • Problem B.1: Solve y′+3xy=x2exy' + \frac{3}{x}y = x^2 e^x, for x>0x > 0

    • First-order linear ODE with P(x)=3xP(x) = \frac{3}{x} and Q(x)=x2exQ(x) = x^2 e^x.
    • Integrating Factor: I.F.=e∫3xdx=e3ln⁡(x)=x3I.F. = e^{\int \frac{3}{x} dx} = e^{3\ln(x)} = x^3.
    • General Solution:         y⋅x3=∫x3⋅(x2ex)dx+C=∫x5exdx+Cy \cdot x^3 = \int x^3 \cdot (x^2 e^x) dx + C = \int x^5 e^x dx + C         Applying integration by parts iteratively:         ∫x5exdx=ex(x5−5x4+20x3−60x2+120x−120)\int x^5 e^x dx = e^x (x^5 - 5x^4 + 20x^3 - 60x^2 + 120x - 120)y(x)=exx3(x5−5x4+20x3−60x2+120x−120)+Cx−3y(x) = \frac{e^x}{x^3} (x^5 - 5x^4 + 20x^3 - 60x^2 + 120x - 120) + C x^{-3}

  • Problem B.2: Solve (D2−4D+3)y=sin⁡(3x)(D^2 - 4D + 3)y = \sin(3x)

    • Auxiliary Equation: m2−4m+3=0  ⟹  m=1,3m^2 - 4m + 3 = 0 \implies m = 1, 3.
    • Complementary Function: yc=C1ex+C2e3xy_c = C_1 e^x + C_2 e^{3x}.
    • Particular Integral: Substitute D2→−32=−9D^2 \to -3^2 = -9.         yp=1−9−4D+3sin⁡(3x)=−1212D+3sin⁡(3x)=−122D−34D2−9sin⁡(3x)y_p = \frac{1}{-9 - 4D + 3} \sin(3x) = -\frac{1}{2} \frac{1}{2D + 3} \sin(3x) = -\frac{1}{2} \frac{2D - 3}{4D^2 - 9} \sin(3x)         Substitute D2→−9D^2 \to -9 in denominator:         yp=−122D−3−36−9sin⁡(3x)=190(6cos⁡(3x)−3sin⁡(3x))=2cos⁡(3x)−sin⁡(3x)30y_p = -\frac{1}{2} \frac{2D - 3}{-36 - 9} \sin(3x) = \frac{1}{90} (6\cos(3x) - 3\sin(3x)) = \frac{2\cos(3x) - \sin(3x)}{30}
    • General Solution: y=C1ex+C2e3x+2cos⁡(3x)−sin⁡(3x)30y = C_1 e^x + C_2 e^{3x} + \frac{2\cos(3x) - \sin(3x)}{30}.

  • Problem B.3: Solve (D2+3D+2)y=sin⁡(x)(D^2 + 3D + 2)y = \sin(x)

    • Auxiliary Equation: m2+3m+2=0  ⟹  m=−1,−2m^2 + 3m + 2 = 0 \implies m = -1, -2.
    • Complementary Function: yc=C1e−x+C2e−2xy_c = C_1 e^{-x} + C_2 e^{-2x}.
    • Particular Integral: Substitute D2→−1D^2 \to -1.         yp=1−1+3D+2sin⁡(x)=3D−19D2−1sin⁡(x)=3D−1−10sin⁡(x)=sin⁡(x)−3cos⁡(x)10y_p = \frac{1}{-1 + 3D + 2} \sin(x) = \frac{3D - 1}{9D^2 - 1} \sin(x) = \frac{3D - 1}{-10} \sin(x) = \frac{\sin(x) - 3\cos(x)}{10}
    • General Solution: y=C1e−x+C2e−2x+sin⁡(x)−3cos⁡(x)10y = C_1 e^{-x} + C_2 e^{-2x} + \frac{\sin(x) - 3\cos(x)}{10}.

  • Problem B.4: Solve d2ydx2+6dydx+9y=3e4x\frac{d^2y}{dx^2} + 6\frac{dy}{dx} + 9y = 3e^{4x}

    • Auxiliary Equation: m2+6m+9=0  ⟹  (m+3)2=0  ⟹  m=−3,−3m^2 + 6m + 9 = 0 \implies (m + 3)^2 = 0 \implies m = -3, -3.
    • Complementary Function: yc=(C1+C2x)e−3xy_c = (C_1 + C_2 x) e^{-3x}.
    • Particular Integral:         yp=1D2+6D+93e4x=3e4x42+6(4)+9=3e4x49y_p = \frac{1}{D^2 + 6D + 9} 3e^{4x} = \frac{3e^{4x}}{4^2 + 6(4) + 9} = \frac{3e^{4x}}{49}
    • General Solution: y=(C1+C2x)e−3x+3e4x49y = (C_1 + C_2 x) e^{-3x} + \frac{3e^{4x}}{49}.

  • Problem B.5: Solve (D2+5D+4)y=e−xsin⁡(2x)(D^2 + 5D + 4)y = e^{-x} \sin(2x)

    • Auxiliary Equation: m2+5m+4=0  ⟹  m=−1,−4m^2 + 5m + 4 = 0 \implies m = -1, -4.
    • Complementary Function: yc=C1e−x+C2e−4xy_c = C_1 e^{-x} + C_2 e^{-4x}.
    • Particular Integral:         yp=e−x1(D−1)2+5(D−1)+4sin⁡(2x)=e−x1D2+3Dsin⁡(2x)y_p = e^{-x} \frac{1}{(D - 1)^2 + 5(D - 1) + 4} \sin(2x) = e^{-x} \frac{1}{D^2 + 3D} \sin(2x)         Substitute D2→−4D^2 \to -4:         yp=e−x13D−4sin⁡(2x)=e−x3D+49D2−16sin⁡(2x)=e−x6cos⁡(2x)+4sin⁡(2x)−52y_p = e^{-x} \frac{1}{3D - 4} \sin(2x) = e^{-x} \frac{3D + 4}{9D^2 - 16} \sin(2x) = e^{-x} \frac{6\cos(2x) + 4\sin(2x)}{-52}yp=−e−x(3cos⁡(2x)+2sin⁡(2x)26)y_p = -e^{-x} \left(\frac{3\cos(2x) + 2\sin(2x)}{26}\right)
    • General Solution: y=C1e−x+C2e−4x−e−x(3cos⁡(2x)+2sin⁡(2x)26)y = C_1 e^{-x} + C_2 e^{-4x} - e^{-x} \left(\frac{3\cos(2x) + 2\sin(2x)}{26}\right).

  • Problem B.6: Find particular integral of (D−3)2y=xe−2x(D - 3)^2 y = x e^{-2x}

    • Particular Integral:         yp=e−2x1(D−2−3)2x=e−2x1(D−5)2x=e−2x125(1−D5)2xy_p = e^{-2x} \frac{1}{(D - 2 - 3)^2} x = e^{-2x} \frac{1}{(D - 5)^2} x = e^{-2x} \frac{1}{25 \left(1 - \frac{D}{5}\right)^2} xyp=e−2x25(1+2D5)x=e−2x25(x+25)=e−2x(5x+2)125y_p = \frac{e^{-2x}}{25} \left(1 + \frac{2D}{5}\right) x = \frac{e^{-2x}}{25} \left(x + \frac{2}{5}\right) = \frac{e^{-2x}(5x + 2)}{125}

  • Problem B.7: Solve (D2+3D+2)y=sin⁡(x)(D^2 + 3D + 2)y = \sin(x), D=ddxD = \frac{d}{dx}

    • Identical to Problem B.3.
    • General Solution: y=C1e−x+C2e−2x+sin⁡(x)−3cos⁡(x)10y = C_1 e^{-x} + C_2 e^{-2x} + \frac{\sin(x) - 3\cos(x)}{10}.

  • Problem B.8: Solve x2d2ydx2+xdydx=0x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} = 0

    • Cauchy-Euler form: Substitute x=ezx = e^z, xD=θx D = \theta, x2D2=θ(θ−1)x^2 D^2 = \theta(\theta - 1).
    • Transformed equation: [θ(θ−1)+θ]y=0  ⟹  θ2y=0[\theta(\theta - 1) + \theta] y = 0 \implies \theta^2 y = 0.
    • Auxiliary Equation: m2=0  ⟹  m=0,0m^2 = 0 \implies m = 0, 0.
    • Solution in zz: y(z)=C1+C2zy(z) = C_1 + C_2 z.
    • General Solution in xx: y(x)=C1+C2ln⁡(x)y(x) = C_1 + C_2 \ln(x).

  • Problem B.9: Solve (D2+4)y=x4(D^2 + 4)y = x^4

    • Identical to Problem A.6.
    • General Solution: y=C1cos⁡(2x)+C2sin⁡(2x)+x44−3x24+38y = C_1 \cos(2x) + C_2 \sin(2x) + \frac{x^4}{4} - \frac{3x^2}{4} + \frac{3}{8}.

  • Problem B.10: Solve d2ydx2+4dydx+4y=e−2x\frac{d^2y}{dx^2} + 4\frac{dy}{dx} + 4y = e^{-2x}

    • Auxiliary Equation: m2+4m+4=0  ⟹  (m+2)2=0  ⟹  m=−2,−2m^2 + 4m + 4 = 0 \implies (m + 2)^2 = 0 \implies m = -2, -2.
    • Complementary Function: yc=(C1+C2x)e−2xy_c = (C_1 + C_2 x) e^{-2x}.
    • Particular Integral:         yp=1(D+2)2e−2x=x212e−2x=x2e−2x2y_p = \frac{1}{(D + 2)^2} e^{-2x} = x^2 \frac{1}{2} e^{-2x} = \frac{x^2 e^{-2x}}{2}
    • General Solution: y=(C1+C2x)e−2x+x2e−2x2y = (C_1 + C_2 x) e^{-2x} + \frac{x^2 e^{-2x}}{2}.

Solutions to Part-C Differential Equations Problems

  • Problem C.11: Solve (D2+6D+8)y=e−2x+cos⁡2(x)(D^2 + 6D + 8)y = e^{-2x} + \cos^2(x)

    • Auxiliary Equation: m2+6m+8=0  ⟹  m=−2,−4m^2 + 6m + 8 = 0 \implies m = -2, -4.
    • Complementary Function: yc=C1e−2x+C2e−4xy_c = C_1 e^{-2x} + C_2 e^{-4x}.
    • Rewrite RHS: e−2x+12+12cos⁡(2x)e^{-2x} + \frac{1}{2} + \frac{1}{2}\cos(2x).
    • Particular Integrals:         yp1=1D2+6D+8e−2x=x12(−2)+6e−2x=xe−2x2y_{p1} = \frac{1}{D^2 + 6D + 8} e^{-2x} = x \frac{1}{2(-2) + 6} e^{-2x} = \frac{x e^{-2x}}{2}yp2=1D2+6D+8(12e0x)=116y_{p2} = \frac{1}{D^2 + 6D + 8} \left(\frac{1}{2} e^{0x}\right) = \frac{1}{16}yp3=121−4+6D+8cos⁡(2x)=143D−29D2−4cos⁡(2x)=143D−2−40cos⁡(2x)=3sin⁡(2x)+cos⁡(2x)80y_{p3} = \frac{1}{2} \frac{1}{-4 + 6D + 8} \cos(2x) = \frac{1}{4} \frac{3D - 2}{9D^2 - 4} \cos(2x) = \frac{1}{4} \frac{3D - 2}{-40} \cos(2x) = \frac{3\sin(2x) + \cos(2x)}{80}
    • General Solution: y=C1e−2x+C2e−4x+xe−2x2+116+3sin⁡(2x)+cos⁡(2x)80y = C_1 e^{-2x} + C_2 e^{-4x} + \frac{x e^{-2x}}{2} + \frac{1}{16} + \frac{3\sin(2x) + \cos(2x)}{80}.

  • Problem C.12: Solve (D2−3D+2)y=e2xsin⁡(x)(D^2 - 3D + 2)y = e^{2x} \sin(x)

    • Auxiliary Equation: m2−3m+2=0  ⟹  m=1,2m^2 - 3m + 2 = 0 \implies m = 1, 2.
    • Complementary Function: yc=C1ex+C2e2xy_c = C_1 e^x + C_2 e^{2x}.
    • Particular Integral:         yp=e2x1(D+2)2−3(D+2)+2sin⁡(x)=e2x1D2+Dsin⁡(x)y_p = e^{2x} \frac{1}{(D + 2)^2 - 3(D + 2) + 2} \sin(x) = e^{2x} \frac{1}{D^2 + D} \sin(x)         Substitute D2→−1D^2 \to -1:         yp=e2x1D−1sin⁡(x)=e2xD+1D2−1sin⁡(x)=−e2x2(cos⁡(x)+sin⁡(x))y_p = e^{2x} \frac{1}{D - 1} \sin(x) = e^{2x} \frac{D + 1}{D^2 - 1} \sin(x) = -\frac{e^{2x}}{2} (\cos(x) + \sin(x))
    • General Solution: y=C1ex+C2e2x−e2x2(cos⁡(x)+sin⁡(x))y = C_1 e^x + C_2 e^{2x} - \frac{e^{2x}}{2} (\cos(x) + \sin(x)).

  • Problem C.13: Solve (D2+3D+2)y=x2+sin⁡(x)(D^2 + 3D + 2)y = x^2 + \sin(x)

    • Complementary Function: yc=C1e−x+C2e−2xy_c = C_1 e^{-x} + C_2 e^{-2x}.
    • Particular Integrals:         yp1=12(1+3D+D22)−1x2=12(1−3D2+7D24)x2=x22−3x2+74y_{p1} = \frac{1}{2} \left(1 + \frac{3D + D^2}{2}\right)^{-1} x^2 = \frac{1}{2} \left(1 - \frac{3D}{2} + \frac{7D^2}{4}\right) x^2 = \frac{x^2}{2} - \frac{3x}{2} + \frac{7}{4}yp2=sin⁡(x)−3cos⁡(x)10y_{p2} = \frac{\sin(x) - 3\cos(x)}{10}
    • General Solution: y=C1e−x+C2e−2x+x22−3x2+74+sin⁡(x)−3cos⁡(x)10y = C_1 e^{-x} + C_2 e^{-2x} + \frac{x^2}{2} - \frac{3x}{2} + \frac{7}{4} + \frac{\sin(x) - 3\cos(x)}{10}.

  • Problem C.14: Solve (D2−2D+1)y=x2+1+sin⁡(2x)(D^2 - 2D + 1)y = x^2 + 1 + \sin(2x)

    • Auxiliary Equation: m2−2m+1=0  ⟹  m=1,1m^2 - 2m + 1 = 0 \implies m = 1, 1.
    • Complementary Function: yc=(C1+C2x)exy_c = (C_1 + C_2 x) e^x.
    • Particular Integrals:         yp1=(1−D)−2(x2+1)=(1+2D+3D2)(x2+1)=x2+4x+7y_{p1} = (1 - D)^{-2} (x^2 + 1) = (1 + 2D + 3D^2)(x^2 + 1) = x^2 + 4x + 7yp2=1−4−2D+1sin⁡(2x)=4cos⁡(2x)−3sin⁡(2x)25y_{p2} = \frac{1}{-4 - 2D + 1} \sin(2x) = \frac{4\cos(2x) - 3\sin(2x)}{25}
    • General Solution: y=(C1+C2x)ex+x2+4x+7+4cos⁡(2x)−3sin⁡(2x)25y = (C_1 + C_2 x) e^x + x^2 + 4x + 7 + \frac{4\cos(2x) - 3\sin(2x)}{25}.

  • Problem C.15: Solve (D2+5D+6)y=x2+4e3x(D^2 + 5D + 6)y = x^2 + 4e^{3x}

    • Auxiliary Equation: m2+5m+6=0  ⟹  m=−2,−3m^2 + 5m + 6 = 0 \implies m = -2, -3.
    • Complementary Function: yc=C1e−2x+C2e−3xy_c = C_1 e^{-2x} + C_2 e^{-3x}.
    • Particular Integrals:         yp1=16(1+5D+D26)−1x2=x26−5x18+19108y_{p1} = \frac{1}{6} \left(1 + \frac{5D + D^2}{6}\right)^{-1} x^2 = \frac{x^2}{6} - \frac{5x}{18} + \frac{19}{108}yp2=4e3x32+5(3)+6=2e3x15y_{p2} = \frac{4e^{3x}}{3^2 + 5(3) + 6} = \frac{2e^{3x}}{15}
    • General Solution: y=C1e−2x+C2e−3x+x26−5x18+19108+2e3x15y = C_1 e^{-2x} + C_2 e^{-3x} + \frac{x^2}{6} - \frac{5x}{18} + \frac{19}{108} + \frac{2e^{3x}}{15}.

  • Problem C.16: Solve (D2+2D+1)y=e3x+sin⁡(2x)(D^2 + 2D + 1)y = e^{3x} + \sin(2x)

    • Complementary Function: yc=(C1+C2x)e−xy_c = (C_1 + C_2 x) e^{-x}.
    • Particular Integrals:         yp1=e3x(3+1)2=e3x16y_{p1} = \frac{e^{3x}}{(3 + 1)^2} = \frac{e^{3x}}{16}yp2=1−4+2D+1sin⁡(2x)=−4cos⁡(2x)+3sin⁡(2x)25y_{p2} = \frac{1}{-4 + 2D + 1} \sin(2x) = -\frac{4\cos(2x) + 3\sin(2x)}{25}
    • General Solution: y=(C1+C2x)e−x+e3x16−4cos⁡(2x)+3sin⁡(2x)25y = (C_1 + C_2 x) e^{-x} + \frac{e^{3x}}{16} - \frac{4\cos(2x) + 3\sin(2x)}{25}.

  • Problem C.17: Solve (D3−3D2+4D−2)y=ex+cos⁡(x)(D^3 - 3D^2 + 4D - 2)y = e^x + \cos(x)

    • Auxiliary Equation: m3−3m2+4m−2=0  ⟹  m=1,1±im^3 - 3m^2 + 4m - 2 = 0 \implies m = 1, 1 \pm i.
    • Complementary Function: yc=C1ex+ex(C2cos⁡(x)+C3sin⁡(x))y_c = C_1 e^x + e^x (C_2 \cos(x) + C_3 \sin(x)).
    • Particular Integrals:         yp1=x13(1)2−6(1)+4ex=xexy_{p1} = x \frac{1}{3(1)^2 - 6(1) + 4} e^x = x e^xyp2=1−D+3+4D−2cos⁡(x)=3D−19D2−1cos⁡(x)=3sin⁡(x)+cos⁡(x)10y_{p2} = \frac{1}{-D + 3 + 4D - 2} \cos(x) = \frac{3D - 1}{9D^2 - 1} \cos(x) = \frac{3\sin(x) + \cos(x)}{10}
    • General Solution: y=C1ex+ex(C2cos⁡(x)+C3sin⁡(x))+xex+3sin⁡(x)+cos⁡(x)10y = C_1 e^x + e^x (C_2 \cos(x) + C_3 \sin(x)) + x e^x + \frac{3\sin(x) + \cos(x)}{10}.

  • Problem C.18: Solve (D2−7D+12)y=5x+cos⁡(2x)(D^2 - 7D + 12)y = 5x + \cos(2x)

    • Auxiliary Equation: m2−7m+12=0  ⟹  m=3,4m^2 - 7m + 12 = 0 \implies m = 3, 4.
    • Complementary Function: yc=C1e3x+C2e4xy_c = C_1 e^{3x} + C_2 e^{4x}.
    • Particular Integrals:         yp1=112(1−7D12)−15x=5x12+35144y_{p1} = \frac{1}{12} \left(1 - \frac{7D}{12}\right)^{-1} 5x = \frac{5x}{12} + \frac{35}{144}yp2=1−4−7D+12cos⁡(2x)=8+7D64−49(−4)cos⁡(2x)=4cos⁡(2x)−7sin⁡(2x)130y_{p2} = \frac{1}{-4 - 7D + 12} \cos(2x) = \frac{8 + 7D}{64 - 49(-4)} \cos(2x) = \frac{4\cos(2x) - 7\sin(2x)}{130}
    • General Solution: y=C1e3x+C2e4x+5x12+35144+4cos⁡(2x)−7sin⁡(2x)130y = C_1 e^{3x} + C_2 e^{4x} + \frac{5x}{12} + \frac{35}{144} + \frac{4\cos(2x) - 7\sin(2x)}{130}.

  • Problem C.19: Solve (D2+6D+9)y=e−2xx3+cos⁡(3x)(D^2 + 6D + 9)y = e^{-2x} x^3 + \cos(3x)

    • Auxiliary Equation: m2+6m+9=0  ⟹  m=−3,−3m^2 + 6m + 9 = 0 \implies m = -3, -3.
    • Complementary Function: yc=(C1+C2x)e−3xy_c = (C_1 + C_2 x) e^{-3x}.
    • Particular Integrals:         yp1=e−2x1(D+1)2x3=e−2x(1−2D+3D2−4D3)x3=e−2x(x3−6x2+18x−24)y_{p1} = e^{-2x} \frac{1}{(D + 1)^2} x^3 = e^{-2x} (1 - 2D + 3D^2 - 4D^3) x^3 = e^{-2x} (x^3 - 6x^2 + 18x - 24)yp2=1−9+6D+9cos⁡(3x)=16Dcos⁡(3x)=sin⁡(3x)18y_{p2} = \frac{1}{-9 + 6D + 9} \cos(3x) = \frac{1}{6D} \cos(3x) = \frac{\sin(3x)}{18}
    • General Solution: y=(C1+C2x)e−3x+e−2x(x3−6x2+18x−24)+sin⁡(3x)18y = (C_1 + C_2 x) e^{-3x} + e^{-2x} (x^3 - 6x^2 + 18x - 24) + \frac{\sin(3x)}{18}.

  • Problem C.20: Solve (x2D2+xD−9)y=5x2(x^2 D^2 + x D - 9)y = \frac{5}{x^2}

    • Cauchy-Euler equation: Let x=ezx = e^z, transformed operator (θ2−9)y=5e−2z(\theta^2 - 9)y = 5e^{-2z}.
    • Auxiliary Equation: m2−9=0  ⟹  m=±3m^2 - 9 = 0 \implies m = \pm 3.
    • Complementary Function: yc=C1x3+C2x−3y_c = C_1 x^3 + C_2 x^{-3}.
    • Particular Integral: yp(z)=5e−2z(−2)2−9=−e−2z=−x−2=−1x2y_p(z) = \frac{5e^{-2z}}{(-2)^2 - 9} = -e^{-2z} = -x^{-2} = -\frac{1}{x^2}.
    • General Solution: y=C1x3+C2x−3−1x2y = C_1 x^3 + C_2 x^{-3} - \frac{1}{x^2}.

  • Problem C.21: Solve (x+2)2d2ydx2−(x+2)dydx+y=3x+4(x + 2)^2 \frac{d^2y}{dx^2} - (x + 2)\frac{dy}{dx} + y = 3x + 4

    • Legendre Linear Equation: Let x+2=ez  ⟹  3x+4=3ez−2x + 2 = e^z \implies 3x + 4 = 3e^z - 2.
    • Operator equation: [θ(θ−1)−θ+1]y=3ez−2  ⟹  (θ−1)2y=3ez−2[\theta(\theta - 1) - \theta + 1] y = 3e^z - 2 \implies (\theta - 1)^2 y = 3e^z - 2.
    • Complementary Function: yc=(C1+C2z)ez=(C1+C2ln⁡(x+2))(x+2)y_c = (C_1 + C_2 z) e^z = (C_1 + C_2 \ln(x + 2)) (x + 2).
    • Particular Integrals:         yp1=3z12(θ−1)ez=32z2ez=32(x+2)[ln⁡(x+2)]2y_{p1} = 3 z \frac{1}{2(\theta - 1)} e^z = \frac{3}{2} z^2 e^z = \frac{3}{2} (x + 2) [\ln(x + 2)]^2yp2=−21=−2y_{p2} = \frac{-2}{1} = -2
    • General Solution: y=(C1+C2ln⁡(x+2))(x+2)+32(x+2)[ln⁡(x+2)]2−2y = (C_1 + C_2 \ln(x + 2))(x + 2) + \frac{3}{2}(x + 2)[\ln(x + 2)]^2 - 2.

  • Problem C.22: Solve (x2D2+xD+1)y=4sin⁡(ln⁡(x))(x^2 D^2 + x D + 1)y = 4 \sin(\ln(x))

    • Cauchy-Euler equation: Let x=ezx = e^z, operator form (θ2+1)y=4sin⁡(z)(\theta^2 + 1)y = 4\sin(z).
    • Complementary Function: yc=C1cos⁡(ln⁡(x))+C2sin⁡(ln⁡(x))y_c = C_1 \cos(\ln(x)) + C_2 \sin(\ln(x)).
    • Particular Integral: yp(z)=4z12θsin⁡(z)=−2zcos⁡(z)=−2ln⁡(x)cos⁡(ln⁡(x))y_p(z) = 4 z \frac{1}{2\theta} \sin(z) = -2z \cos(z) = -2\ln(x) \cos(\ln(x)).
    • General Solution: y=C1cos⁡(ln⁡(x))+C2sin⁡(ln⁡(x))−2ln⁡(x)cos⁡(ln⁡(x))y = C_1 \cos(\ln(x)) + C_2 \sin(\ln(x)) - 2\ln(x) \cos(\ln(x)).

  • Problem C.23: Solve (x2D2−2xD−4)y=32(ln⁡(x))2(x^2 D^2 - 2x D - 4)y = 32(\ln(x))^2

    • Cauchy-Euler equation: Let x=ezx = e^z, operator form (θ2−3θ−4)y=32z2(\theta^2 - 3\theta - 4)y = 32 z^2.
    • Auxiliary Equation: m2−3m−4=0  ⟹  m=4,−1m^2 - 3m - 4 = 0 \implies m = 4, -1.
    • Complementary Function: yc=C1x4+C2x−1y_c = C_1 x^4 + C_2 x^{-1}.
    • Particular Integral:         yp(z)=−8(1−3θ−θ24)−1z2=−8z2+12z−13y_p(z) = -8 \left(1 - \frac{3\theta - \theta^2}{4}\right)^{-1} z^2 = -8 z^2 + 12 z - 13yp(x)=−8(ln⁡(x))2+12ln⁡(x)−13y_p(x) = -8(\ln(x))^2 + 12\ln(x) - 13
    • General Solution: y=C1x4+C2x−1−8(ln⁡(x))2+12ln⁡(x)−13y = C_1 x^4 + C_2 x^{-1} - 8(\ln(x))^2 + 12\ln(x) - 13.

  • Problem C.24: Solve (x2D2−xD+1)y=(ln⁡(x)x)2(x^2 D^2 - x D + 1)y = \left(\frac{\ln(x)}{x}\right)^2

    • Cauchy-Euler equation: Let x=ezx = e^z, operator form (θ−1)2y=z2e−2z(\theta - 1)^2 y = z^2 e^{-2z}.
    • Complementary Function: yc=(C1+C2ln⁡(x))xy_c = (C_1 + C_2 \ln(x)) x.
    • Particular Integral:         yp(z)=e−2z1(θ−3)2z2=e−2z9(1−θ3)−2z2=e−2z27(3z2+4z+2)y_p(z) = e^{-2z} \frac{1}{(\theta - 3)^2} z^2 = \frac{e^{-2z}}{9} \left(1 - \frac{\theta}{3}\right)^{-2} z^2 = \frac{e^{-2z}}{27} (3z^2 + 4z + 2)yp(x)=3(ln⁡(x))2+4ln⁡(x)+227x2y_p(x) = \frac{3(\ln(x))^2 + 4\ln(x) + 2}{27x^2}
    • General Solution: y=(C1+C2ln⁡(x))x+3(ln⁡(x))2+4ln⁡(x)+227x2y = (C_1 + C_2 \ln(x)) x + \frac{3(\ln(x))^2 + 4\ln(x) + 2}{27x^2}.

  • Problem C.25: Solve (x2D2−3xD+4)y=x2(x^2 D^2 - 3x D + 4)y = x^2

    • Cauchy-Euler equation: Let x=ezx = e^z, operator form (θ−2)2y=e2z(\theta - 2)^2 y = e^{2z}.
    • Complementary Function: yc=(C1+C2ln⁡(x))x2y_c = (C_1 + C_2 \ln(x)) x^2.
    • Particular Integral: yp(z)=z2e2z2=x2(ln⁡(x))22y_p(z) = \frac{z^2 e^{2z}}{2} = \frac{x^2 (\ln(x))^2}{2}.
    • General Solution: y=(C1+C2ln⁡(x))x2+x2(ln⁡(x))22y = (C_1 + C_2 \ln(x)) x^2 + \frac{x^2 (\ln(x))^2}{2}.

  • Problem C.26: Solve d2ydx2+1xdydx−12ln⁡(x)x2y=0\frac{d^2y}{dx^2} + \frac{1}{x}\frac{dy}{dx} - \frac{12\ln(x)}{x^2} y = 0

    • Multiply by x2x^2: x2d2ydx2+xdydx−12ln⁡(x)y=0x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} - 12\ln(x) y = 0.
    • Transforming variable via x=ezx = e^z yields d2ydz2−12zy=0\frac{d^2y}{dz^2} - 12z y = 0.

  • Problem C.27: Solve Euler's equation x2d2ydx2−7xdydx+12y=x2x^2 \frac{d^2y}{dx^2} - 7x \frac{dy}{dx} + 12y = x^2

    • Cauchy-Euler equation: Let x=ezx = e^z, operator form (θ2−8θ+12)y=e2z(\theta^2 - 8\theta + 12)y = e^{2z}.
    • Auxiliary Equation: m2−8m+12=0  ⟹  m=2,6m^2 - 8m + 12 = 0 \implies m = 2, 6.
    • Complementary Function: yc=C1x2+C2x6y_c = C_1 x^2 + C_2 x^6.
    • Particular Integral: yp(z)=z12(2)−8e2z=−ze2z4=−x2ln⁡(x)4y_p(z) = z \frac{1}{2(2) - 8} e^{2z} = -\frac{z e^{2z}}{4} = -\frac{x^2 \ln(x)}{4}.
    • General Solution: y=C1x2+C2x6−x2ln⁡(x)4y = C_1 x^2 + C_2 x^6 - \frac{x^2 \ln(x)}{4}.

  • Problem C.28: Solve x2d2ydx2+4xdydx+2y=xln⁡(x)x^2 \frac{d^2y}{dx^2} + 4x \frac{dy}{dx} + 2y = x \ln(x)

    • Cauchy-Euler equation: Let x=ezx = e^z, operator form (θ2+3θ+2)y=zez(\theta^2 + 3\theta + 2)y = z e^z.
    • Auxiliary Equation: m2+3m+2=0  ⟹  m=−1,−2m^2 + 3m + 2 = 0 \implies m = -1, -2.
    • Complementary Function: yc=C1x−1+C2x−2y_c = C_1 x^{-1} + C_2 x^{-2}.
    • Particular Integral:         yp(z)=ez1θ2+5θ+6z=ez6(1−5θ6)z=ez6(z−56)y_p(z) = e^z \frac{1}{\theta^2 + 5\theta + 6} z = \frac{e^z}{6} \left(1 - \frac{5\theta}{6}\right) z = \frac{e^z}{6} \left(z - \frac{5}{6}\right)yp(x)=xln⁡(x)6−5x36y_p(x) = \frac{x \ln(x)}{6} - \frac{5x}{36}
    • General Solution: y=C1x−1+C2x−2+xln⁡(x)6−5x36y = C_1 x^{-1} + C_2 x^{-2} + \frac{x \ln(x)}{6} - \frac{5x}{36}.

  • Problem C.29: Solve by Variation of Parameters: d2ydx2+y=tan⁡(x)\frac{d^2y}{dx^2} + y = \tan(x)

    • Complementary solution: yc=C1cos⁡(x)+C2sin⁡(x)y_c = C_1 \cos(x) + C_2 \sin(x), with y1=cos⁡(x),y2=sin⁡(x)y_1 = \cos(x), y_2 = \sin(x).
    • Wronskian: W(y1,y2)=cos⁡2(x)+sin⁡2(x)=1W(y_1, y_2) = \cos^2(x) + \sin^2(x) = 1.
    • Parameters:         u(x)=−∫sin⁡(x)tan⁡(x)1dx=−∫(sec⁡(x)−cos⁡(x))dx=−ln⁡∣sec⁡(x)+tan⁡(x)∣+sin⁡(x)u(x) = -\int \frac{\sin(x)\tan(x)}{1} dx = -\int (\sec(x) - \cos(x)) dx = -\ln|\sec(x) + \tan(x)| + \sin(x)v(x)=∫cos⁡(x)tan⁡(x)1dx=−cos⁡(x)v(x) = \int \frac{\cos(x)\tan(x)}{1} dx = -\cos(x)
    • Particular Integral:         yp=−cos⁡(x)ln⁡∣sec⁡(x)+tan⁡(x)∣+sin⁡(x)cos⁡(x)−sin⁡(x)cos⁡(x)=−cos⁡(x)ln⁡∣sec⁡(x)+tan⁡(x)∣y_p = -\cos(x) \ln|\sec(x) + \tan(x)| + \sin(x)\cos(x) - \sin(x)\cos(x) = -\cos(x) \ln|\sec(x) + \tan(x)|
    • General Solution: y=C1cos⁡(x)+C2sin⁡(x)−cos⁡(x)ln⁡∣sec⁡(x)+tan⁡(x)∣y = C_1 \cos(x) + C_2 \sin(x) - \cos(x) \ln|\sec(x) + \tan(x)|.

  • Problem C.30: Solve (D2+4)y=4tan⁡(2x)(D^2 + 4)y = 4 \tan(2x) using Variation of Parameters

    • Complementary solution: yc=C1cos⁡(2x)+C2sin⁡(2x)y_c = C_1 \cos(2x) + C_2 \sin(2x), with y1=cos⁡(2x),y2=sin⁡(2x)y_1 = \cos(2x), y_2 = \sin(2x).
    • Wronskian: W=2W = 2.
    • Parameters:         u(x)=−∫sin⁡(2x)⋅4tan⁡(2x)2dx=−ln⁡∣sec⁡(2x)+tan⁡(2x)∣+sin⁡(2x)u(x) = -\int \frac{\sin(2x) \cdot 4\tan(2x)}{2} dx = -\ln|\sec(2x) + \tan(2x)| + \sin(2x)v(x)=∫cos⁡(2x)⋅4tan⁡(2x)2dx=−cos⁡(2x)v(x) = \int \frac{\cos(2x) \cdot 4\tan(2x)}{2} dx = -\cos(2x)
    • Particular Integral: yp=−cos⁡(2x)ln⁡∣sec⁡(2x)+tan⁡(2x)∣y_p = -\cos(2x) \ln|\sec(2x) + \tan(2x)|.
    • General Solution: y=C1cos⁡(2x)+C2sin⁡(2x)−cos⁡(2x)ln⁡∣sec⁡(2x)+tan⁡(2x)∣y = C_1 \cos(2x) + C_2 \sin(2x) - \cos(2x) \ln|\sec(2x) + \tan(2x)|.

  • Problem C.31: Solve by Variation of Parameters: d2ydx2+y=csc⁡(x)\frac{d^2y}{dx^2} + y = \csc(x)

    • Complementary solution: yc=C1cos⁡(x)+C2sin⁡(x)y_c = C_1 \cos(x) + C_2 \sin(x), W=1W = 1.
    • Parameters:         u(x)=−∫sin⁡(x)csc⁡(x)dx=−xu(x) = -\int \sin(x)\csc(x) dx = -xv(x)=∫cos⁡(x)csc⁡(x)dx=ln⁡∣sin⁡(x)∣v(x) = \int \cos(x)\csc(x) dx = \ln|\sin(x)|
    • Particular Integral: yp=−xcos⁡(x)+sin⁡(x)ln⁡∣sin⁡(x)∣y_p = -x \cos(x) + \sin(x) \ln|\sin(x)|.
    • General Solution: y=C1cos⁡(x)+C2sin⁡(x)−xcos⁡(x)+sin⁡(x)ln⁡∣sin⁡(x)∣y = C_1 \cos(x) + C_2 \sin(x) - x \cos(x) + \sin(x) \ln|\sin(x)|.

  • Problem C.32: Solve d2ydx2+y=cot⁡(x)\frac{d^2y}{dx^2} + y = \cot(x) using Variation of Parameters

    • Complementary solution: yc=C1cos⁡(x)+C2sin⁡(x)y_c = C_1 \cos(x) + C_2 \sin(x), W=1W = 1.
    • Parameters:         u(x)=−∫sin⁡(x)cot⁡(x)dx=−sin⁡(x)u(x) = -\int \sin(x)\cot(x) dx = -\sin(x)v(x)=∫cos⁡2(x)sin⁡(x)dx=ln⁡∣csc⁡(x)−cot⁡(x)∣+cos⁡(x)v(x) = \int \frac{\cos^2(x)}{\sin(x)} dx = \ln|\csc(x) - \cot(x)| + \cos(x)
    • Particular Integral: yp=sin⁡(x)ln⁡∣csc⁡(x)−cot⁡(x)∣y_p = \sin(x) \ln|\csc(x) - \cot(x)|.
    • General Solution: y=C1cos⁡(x)+C2sin⁡(x)+sin⁡(x)ln⁡∣csc⁡(x)−cot⁡(x)∣y = C_1 \cos(x) + C_2 \sin(x) + \sin(x) \ln|\csc(x) - \cot(x)|.

  • Problem C.33: Solve d2ydx2+4y=sec⁡(2x)\frac{d^2y}{dx^2} + 4y = \sec(2x) using Variation of Parameters

    • Complementary solution: yc=C1cos⁡(2x)+C2sin⁡(2x)y_c = C_1 \cos(2x) + C_2 \sin(2x), W=2W = 2.
    • Parameters:         u(x)=−12∫tan⁡(2x)dx=−14ln⁡∣sec⁡(2x)∣u(x) = -\frac{1}{2} \int \tan(2x) dx = -\frac{1}{4} \ln|\sec(2x)|v(x)=12∫1dx=x2v(x) = \frac{1}{2} \int 1 dx = \frac{x}{2}
    • Particular Integral: yp=−14cos⁡(2x)ln⁡∣sec⁡(2x)∣+x2sin⁡(2x)y_p = -\frac{1}{4} \cos(2x) \ln|\sec(2x)| + \frac{x}{2} \sin(2x).
    • General Solution: y=C1cos⁡(2x)+C2sin⁡(2x)−14cos⁡(2x)ln⁡∣sec⁡(2x)∣+x2sin⁡(2x)y = C_1 \cos(2x) + C_2 \sin(2x) - \frac{1}{4} \cos(2x) \ln|\sec(2x)| + \frac{x}{2} \sin(2x).

  • Problem C.34: Solve d2ydx2+a2y=tan⁡(ax)\frac{d^2y}{dx^2} + a^2 y = \tan(ax) using Variation of Parameters

    • Complementary solution: yc=C1cos⁡(ax)+C2sin⁡(ax)y_c = C_1 \cos(ax) + C_2 \sin(ax), W=aW = a.
    • Parameters:         u(x)=−1a∫sin⁡2(ax)cos⁡(ax)dx=−1a2ln⁡∣sec⁡(ax)+tan⁡(ax)∣+1a2sin⁡(ax)u(x) = -\frac{1}{a} \int \frac{\sin^2(ax)}{\cos(ax)} dx = -\frac{1}{a^2} \ln|\sec(ax) + \tan(ax)| + \frac{1}{a^2} \sin(ax)v(x)=1a∫sin⁡(ax)dx=−1a2cos⁡(ax)v(x) = \frac{1}{a} \int \sin(ax) dx = -\frac{1}{a^2} \cos(ax)
    • Particular Integral: yp=−1a2cos⁡(ax)ln⁡∣sec⁡(ax)+tan⁡(ax)∣y_p = -\frac{1}{a^2} \cos(ax) \ln|\sec(ax) + \tan(ax)|.
    • General Solution: y=C1cos⁡(ax)+C2sin⁡(ax)−1a2cos⁡(ax)ln⁡∣sec⁡(ax)+tan⁡(ax)∣y = C_1 \cos(ax) + C_2 \sin(ax) - \frac{1}{a^2} \cos(ax) \ln|\sec(ax) + \tan(ax)|.