Particular Solutions to First-Order Differential Equations: Exercises 11-14 Problem Fragment Analysis: Top Page Entry Original Transcript Text: e tan y dx + (1-e)Context: This line appears at the beginning of the provided material. It is an incomplete fragment of a first-order ordinary differential equation.Potential Full Form: Given the standard formats in mathematics textbooks, this fragment likely represents an entry such as:
e x tan ( y ) d x + ( 1 − e x ) sec 2 ( y ) d y = 0 e^x \tan(y)\,dx + (1-e^x) \sec^2(y)\,dy = 0 e x tan ( y ) d x + ( 1 − e x ) sec 2 ( y ) d y = 0 Components Identified: e e e e x e^x e x tan ( y ) \tan(y) tan ( y ) y y y d x dx d x x x x ( 1 − e ) (1-e) ( 1 − e ) Exercise 11: Rational Function Integration and Particular Solutions Differential Equation: ( x 3 + x 2 + x + 1 ) d y d x = 2 x 2 + x (x^3+x^2+x+1) \frac{dy}{dx} = 2x^2 + x ( x 3 + x 2 + x + 1 ) d x d y = 2 x 2 + x Initial Condition: y = 1 y = 1 y = 1 x = 0 x = 0 x = 0 Methodology: Separation of Variables Rearrange the equation to isolate d y d x \frac{dy}{dx} d x d y d y d x = 2 x 2 + x x 3 + x 2 + x + 1 \frac{dy}{dx} = \frac{2x^2 + x}{x^3 + x^2 + x + 1} d x d y = x 3 + x 2 + x + 1 2 x 2 + x Separate the variables and integrate both sides:
∫ d y = ∫ 2 x 2 + x x 3 + x 2 + x + 1 d x \int dy = \int \frac{2x^2 + x}{x^3 + x^2 + x + 1}\,dx ∫ d y = ∫ x 3 + x 2 + x + 1 2 x 2 + x d x Step-by-Step Solving: Factoring the Denominator: The polynomial x 3 + x 2 + x + 1 x^3 + x^2 + x + 1 x 3 + x 2 + x + 1 x 2 ( x + 1 ) + 1 ( x + 1 ) = ( x + 1 ) ( x 2 + 1 ) x^2(x + 1) + 1(x + 1) = (x + 1)(x^2 + 1) x 2 ( x + 1 ) + 1 ( x + 1 ) = ( x + 1 ) ( x 2 + 1 ) Partial Fraction Decomposition: 2 x 2 + x ( x + 1 ) ( x 2 + 1 ) = A x + 1 + B x + C x 2 + 1 \frac{2x^2 + x}{(x + 1)(x^2 + 1)} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 + 1} ( x + 1 ) ( x 2 + 1 ) 2 x 2 + x = x + 1 A + x 2 + 1 B x + C 2 x 2 + x = A ( x 2 + 1 ) + ( B x + C ) ( x + 1 ) 2x^2 + x = A(x^2 + 1) + (Bx + C)(x + 1) 2 x 2 + x = A ( x 2 + 1 ) + ( B x + C ) ( x + 1 ) Setting x = − 1 x = -1 x = − 1 2 − 1 = A ( 1 + 1 ) → 1 = 2 A → A = 1 2 2 - 1 = A(1 + 1) \rightarrow 1 = 2A \rightarrow A = \frac{1}{2} 2 − 1 = A ( 1 + 1 ) → 1 = 2 A → A = 2 1 Comparing x 2 x^2 x 2 2 = A + B → 2 = 1 2 + B → B = 3 2 2 = A + B \rightarrow 2 = \frac{1}{2} + B \rightarrow B = \frac{3}{2} 2 = A + B → 2 = 2 1 + B → B = 2 3 Comparing constant terms: 0 = A + C → 0 = 1 2 + C → C = − 1 2 0 = A + C \rightarrow 0 = \frac{1}{2} + C \rightarrow C = -\frac{1}{2} 0 = A + C → 0 = 2 1 + C → C = − 2 1 Integrating the Terms: y = ∫ ( 1 2 ( x + 1 ) + 3 2 x − 1 2 x 2 + 1 ) d x y = \int \left( \frac{1}{2(x+1)} + \frac{\frac{3}{2}x - \frac{1}{2}}{x^2+1} \right)\,dx y = ∫ ( 2 ( x + 1 ) 1 + x 2 + 1 2 3 x − 2 1 ) d x y = 1 2 ln ∣ x + 1 ∣ + 3 2 ∫ x x 2 + 1 d x − 1 2 ∫ 1 x 2 + 1 d x y = \frac{1}{2} \ln|x+1| + \frac{3}{2} \int \frac{x}{x^2+1}\,dx - \frac{1}{2} \int \frac{1}{x^2+1}\,dx y = 2 1 ln ∣ x + 1∣ + 2 3 ∫ x 2 + 1 x d x − 2 1 ∫ x 2 + 1 1 d x y = 1 2 ln ∣ x + 1 ∣ + 3 4 ln ( x 2 + 1 ) − 1 2 tan − 1 ( x ) + C y = \frac{1}{2} \ln|x+1| + \frac{3}{4} \ln(x^2+1) - \frac{1}{2} \tan^{-1}(x) + C y = 2 1 ln ∣ x + 1∣ + 4 3 ln ( x 2 + 1 ) − 2 1 tan − 1 ( x ) + C Finding the Particular Solution: Substitute x = 0 x = 0 x = 0 y = 1 y = 1 y = 1 1 = 1 2 ln ( 1 ) + 3 4 ln ( 1 ) − 1 2 tan − 1 ( 0 ) + C 1 = \frac{1}{2} \ln(1) + \frac{3}{4} \ln(1) - \frac{1}{2} \tan^{-1}(0) + C 1 = 2 1 ln ( 1 ) + 4 3 ln ( 1 ) − 2 1 tan − 1 ( 0 ) + C 1 = 0 + 0 − 0 + C → C = 1 1 = 0 + 0 - 0 + C \rightarrow C = 1 1 = 0 + 0 − 0 + C → C = 1 Final Solution: y = 1 2 ln ∣ x + 1 ∣ + 3 4 ln ( x 2 + 1 ) − 1 2 tan − 1 ( x ) + 1 y = \frac{1}{2} \ln|x+1| + \frac{3}{4} \ln(x^2+1) - \frac{1}{2} \tan^{-1}(x) + 1 y = 2 1 ln ∣ x + 1∣ + 4 3 ln ( x 2 + 1 ) − 2 1 tan − 1 ( x ) + 1 Differential Equation: x ( x 2 − 1 ) d y d x = 1 x(x^2-1) \frac{dy}{dx} = 1 x ( x 2 − 1 ) d x d y = 1 Initial Condition: y = 0 y = 0 y = 0 x = 2 x = 2 x = 2 Methodology: Separation of Variables Separate the differentials:
d y = 1 x ( x + 1 ) ( x − 1 ) d x dy = \frac{1}{x(x+1)(x-1)}\,dx d y = x ( x + 1 ) ( x − 1 ) 1 d x Step-by-Step Solving: Partial Fraction Decomposition: 1 x ( x − 1 ) ( x + 1 ) = A x + B x − 1 + C x + 1 \frac{1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1} x ( x − 1 ) ( x + 1 ) 1 = x A + x − 1 B + x + 1 C Using the cover-up method:A = 1 0 2 − 1 = − 1 A = \frac{1}{0^2 - 1} = -1 A = 0 2 − 1 1 = − 1 B = 1 1 ( 1 + 1 ) = 1 2 B = \frac{1}{1(1+1)} = \frac{1}{2} B = 1 ( 1 + 1 ) 1 = 2 1 C = 1 − 1 ( − 1 − 1 ) = 1 2 C = \frac{1}{-1(-1-1)} = \frac{1}{2} C = − 1 ( − 1 − 1 ) 1 = 2 1 Integration: y = ∫ ( − 1 x + 1 2 ( x − 1 ) + 1 2 ( x + 1 ) ) d x y = \int \left( -\frac{1}{x} + \frac{1}{2(x-1)} + \frac{1}{2(x+1)} \right)\,dx y = ∫ ( − x 1 + 2 ( x − 1 ) 1 + 2 ( x + 1 ) 1 ) d x y = − ln ∣ x ∣ + 1 2 ln ∣ x − 1 ∣ + 1 2 ln ∣ x + 1 ∣ + C y = -\ln|x| + \frac{1}{2} \ln|x-1| + \frac{1}{2} \ln|x+1| + C y = − ln ∣ x ∣ + 2 1 ln ∣ x − 1∣ + 2 1 ln ∣ x + 1∣ + C y = ln ( x 2 − 1 x ) + C y = \ln\left( \frac{\sqrt{x^2-1}}{x} \right) + C y = ln ( x x 2 − 1 ) + C Finding the Particular Solution: Substitute x = 2 , y = 0 x = 2, y = 0 x = 2 , y = 0 0 = ln ( 4 − 1 2 ) + C 0 = \ln\left( \frac{\sqrt{4-1}}{2} \right) + C 0 = ln ( 2 4 − 1 ) + C C = − ln ( 3 2 ) C = -\ln\left( \frac{\sqrt{3}}{2} \right) C = − ln ( 2 3 ) Final Solution: y = ln ( x 2 − 1 x ) − ln ( 3 2 ) = ln ( 2 x 2 − 1 3 x ) y = \ln\left( \frac{\sqrt{x^2-1}}{x} \right) - \ln\left( \frac{\sqrt{3}}{2} \right) = \ln\left( \frac{2\sqrt{x^2-1}}{\sqrt{3}x} \right) y = ln ( x x 2 − 1 ) − ln ( 2 3 ) = ln ( 3 x 2 x 2 − 1 ) Exercise 13: Solution for Constant Derivatives Differential Equation: cos ( d y d x ) = a \cos\left(\frac{dy}{dx}\right) = a cos ( d x d y ) = a Note: It is specified that a ∈ R a \in \mathbb{R} a ∈ R Initial Condition: y = 1 y = 1 y = 1 x = 0 x = 0 x = 0 Methodology: The equation implies the derivative is a constant value determined by the inverse cosine of a a a Express the derivative: d y d x = cos − 1 ( a ) \frac{dy}{dx} = \cos^{-1}(a) d x d y = cos − 1 ( a ) Step-by-Step Solving: Separate variables:
d y = cos − 1 ( a ) d x dy = \cos^{-1}(a)\,dx d y = cos − 1 ( a ) d x Integrate both sides:
∫ d y = cos − 1 ( a ) ∫ d x \int dy = \cos^{-1}(a) \int dx ∫ d y = cos − 1 ( a ) ∫ d x y = x cos − 1 ( a ) + C y = x \cos^{-1}(a) + C y = x cos − 1 ( a ) + C Finding the Particular Solution: Substitute x = 0 , y = 1 x = 0, y = 1 x = 0 , y = 1 1 = ( 0 ) cos − 1 ( a ) + C → C = 1 1 = (0) \cos^{-1}(a) + C \rightarrow C = 1 1 = ( 0 ) cos − 1 ( a ) + C → C = 1 Final Solution: y = x cos − 1 ( a ) + 1 y = x \cos^{-1}(a) + 1 y = x cos − 1 ( a ) + 1 Exercise 14: Integrating Trig Functions with Separation of Variables Differential Equation: d y d x = y tan ( x ) \frac{dy}{dx} = y \tan(x) d x d y = y tan ( x ) Initial Condition: y = 1 y = 1 y = 1 x = 0 x = 0 x = 0 Methodology: Separation of Variables Group terms with respect to their variables:
1 y d y = tan ( x ) d x \frac{1}{y}\,dy = \tan(x)\,dx y 1 d y = tan ( x ) d x Step-by-Step Solving: Integrate both sides:
∫ 1 y d y = ∫ tan ( x ) d x \int \frac{1}{y}\,dy = \int \tan(x)\,dx ∫ y 1 d y = ∫ tan ( x ) d x Recall the integral of the tangent function:
ln ∣ y ∣ = ln ∣ sec ( x ) ∣ + C \ln|y| = \ln|\sec(x)| + C ln ∣ y ∣ = ln ∣ sec ( x ) ∣ + C Transform the logarithmic equation:
y = e ln ∣ sec ( x ) ∣ + C = e C ⋅ ∣ sec ( x ) ∣ y = e^{\ln|\sec(x)| + C} = e^C \cdot |\sec(x)| y = e l n ∣ s e c ( x ) ∣ + C = e C ⋅ ∣ sec ( x ) ∣ A = e C A = e^C A = e C y = A sec ( x ) y = A \sec(x) y = A sec ( x ) Finding the Particular Solution: Substitute x = 0 , y = 1 x = 0, y = 1 x = 0 , y = 1 1 = A sec ( 0 ) 1 = A \sec(0) 1 = A sec ( 0 ) sec ( 0 ) = 1 cos ( 0 ) = 1 \sec(0) = \frac{1}{\cos(0)} = 1 sec ( 0 ) = c o s ( 0 ) 1 = 1 1 = A ( 1 ) → A = 1 1 = A(1) \rightarrow A = 1 1 = A ( 1 ) → A = 1 Final Solution: y = sec ( x ) y = \sec(x) y = sec ( x )