integrals, derivatives and differential equations

\frac{dy}{\differentialD x}sin(x)

cos(x)

$$\frac{dy}{dx}cos(x)$$

-sin(x)

$$\frac{dy}{dx} tan(x)$$

$$sec² (x)$$

$$\frac{dy}{dx}sec(x)$$

$$sec(x)tan(x)$$

$$\frac{dy}{dx} cosecant(x)$$

$$-csc(x)cot(x)$$

$$\frac{dy}{dx} cot(x)$$

$$-csc² (x)$$

$$\frac{dy}{dx} e^{ax}$$

$$ae^{ax}$$

$$\frac{dy}{dx} a^x$$

what is the theory?

$$a^x ln(a)$$

can be written as $e^{xln(a)}$

$$\frac{dy}{dx} a^{bx+c}$$

$$b ln(a) a^{bx+c}$$

$$\frac{dy}{dx} : ln(x)$$

$$\frac{1}{x}$$

$$\frac{dy}{dx} ln(x^a)$$

$$\frac{1}{u}*u’ → \frac{1}{x^a}*ax^{a-1} → x^{-a}ax^{a-1} → ax^{-1} → \frac{a}{x}$$

$$\frac{dy}{dx} log_a (u)$$

$$\frac{u’}{u ln(a)}$$

integral of (ax+b)ⁿ

$$\frac{1}{a}*\frac{(ax+b)^{n+1}}{n+1}$$

integral of $\frac{1}{(ax+b)}$

$$\frac{ln(ax+b)}{a}$$

integral of $e^{ax+b}$

$$\frac{e^{ax+b}}{a}$$

integral of $a^{bx+c}$

theory?

just like the derivative you can write the equation as $e^{bx ln(a) + c ln(a)}$

you can then integrate it the same as an exponential

$$\frac{a^{bx+c}}{b ln(a)}$$

integral of ln(X) theory?

integrate by parts:

u = ln(x)

du = $\frac{1}{x}$

v = x

dv = dx

$$x ln(x)-x$$

integral of $cos$

sin(x)

integral of sin

-cos

derivative of $cos^{-1}$

$$-\frac{1}{\sqrt{1-x²}}$$

derivative of $sin^{-1}$

$$\frac{1}{\sqrt{1-x²}}$$

derivative of $tan^{-1}$

$$\frac{1}{x²+1}$$

derivative of $cot^{-1}$

$$-\frac{1}{x²+1}$$

derivative of %%sec^{-1}$$

$$\frac{1}{|X|\sqrt{X²-1}}$$

derivative of $csc^{-1}$

$$-\frac{1}{|X|\sqrt{X²-1}}$$

linear equation

$$\frac{dy}{dy} + Py = Q$$

solved using $e^{\int p}dx$

differential is not multiplied by anything containing y

separable equation

$$dy/dx +ay = 0$$

can seperate the x and y variables

Homogenous equation

$$\frac{dy}{dx} * P = Q$$

all variables are to the same order

solved by substituting y = vx

exact equation

$$f(x,y) + g(x,y) \frac{dy}{dx} = m(x)$$

can have an equation only containing x terms on the right

solved by finding if h(x,y) is real

$$\frac{d²y}{dx²} = x$$

solved by integrating twice remember that it produces 2 constants

$$\frac{d²y}{dx²} = y$$

mutiply both sides by $2\frac{d²y}{dx²}$ remember that $2(dy/dx)\frac{d²y}{dx²} = \frac{d}{dx} (\frac{dy}{dx})^2$

$$\frac{d²y}{dx²} +b(\frac{dy}{dx})+y= 0$$

find the auxillary equation then the complimentary equation

1 root = $e^{mx} (Ax+B)$

2 real roots = $Ae^{m_1x} + Be^{m_2x}$

no real roots = $Ae^{(P+q)x} + Be^{(P-q)x} or e^{Px} ( Acos(qx) + Bsin(qx))$

$$\frac{d²y}{dx²} +b(\frac{dy}{dx})+y= polynomial$$

particular integral: $a_n x^n + a_{n-1} x^{n-1}+ …+ a_o$

integrate once and twice

substitute back into equation and solve for a

to find general solution mash everything up with the complimentary equation

$$\frac{d²y}{dx²} +b(\frac{dy}{dx})+y= exponential$$

$y=Pe^{mx}$ if the complimentary equation appears in the original equation use $y = Px^n e^{mx}$ where n is the number of times the solution appears&

$$\frac{d²y}{dx²} +b(\frac{dy}{dx})+y= trig function$$

$y=Pcos(rx)+Qsin(rx)$ where r is the value that appears next to x in your actual equation

if the solution to the complimentary equation appears in the actual equation use $x^n ( Pcos(rx) + Qsin(rx))$