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Core Definitions

Ordinary differential equation (ODE): Differential equation with only one independent variable.
Order: Highest order derivative present.
Degree: Power of highest order derivative after equation made free from radicals/fractions in derivatives.
Linear differential equation: $\frac{dy}{dx} + P(x)y = Q(x)$
Bernoulli equation: $\frac{dy}{dx} + P(x)y = Q(x)y^n,\quad n \ne 0,1$

Fast Classification

$y' = e^x$ : order $=1$ , degree $=1$
$(y')^2 + y' - \tan^2 y = 0$ : order $=1$ , degree $=2$
$y''' + y'' - y' + y = x\ln x$ : order $=3$ , degree $=1$
$(y'')^3 + 6y' = 0$ : order $=2$ , degree $=3$

Forming DE

If solution has 1 constant: differentiate once.
If solution has 2 constants: differentiate twice.
If solution has 3 constants: differentiate thrice.
Eliminate arbitrary constants from original + derived equations.

Examples:

$y = Ae^{2x} + Be^x + C \Rightarrow y''' - 3y'' + 2y' = 0$
$y = e^x(A\cos x + B\sin x) \Rightarrow y'' - 2y' + 2y = 0$

Separation of Variables

Goal:

\frac{dy}{dx} = f(x)g(y) \Rightarrow \frac{dy}{g(y)} = f(x)\,dx

Useful substitutions from exam:

If $\sin^{-1}(y') = x+y$ , take $\sin$ both sides, then set $z=x+y$
If $y'=(4x+y+1)^2$ , set $z=4x+y+1$

Results to remember:

$\sin^{-1}(y') = x+y \Rightarrow \tan(x+y)-\sec(x+y)=x+C$
$y'=(4x+y+1)^2 \Rightarrow \frac12 \tan^{-1}\!\left(\frac{4x+y+1}{2}\right)=x+C$

Exact DE

For

M(x,y)\,dx + N(x,y)\,dy = 0

exact iff

\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}

Solution pattern:

Integrate $M$ wrt $x$
Add function of $y$
Match with $N$

Exam results:

$(1+e^{x/y})dx + e^{x/y}(1-x/y)dy=0$ $x + ye^{x/y} = C$
$(x^2y-2xy^2)dx-(x^3-3x^2y)dy=0$ after I.F. $\frac{x}{y} + \ln\!\left(\frac{y^3}{x^2}\right)=C$

Linear DE Shortcut

Standard form:

\frac{dy}{dx}+P(x)y=Q(x)

Integrating factor:

\text{I.F.}=e^{\int P(x)\,dx}

Solution:

y\cdot \text{I.F.}=\int Q(x)\,\text{I.F.}\,dx + C

Exam case:

(1+x^2)\frac{dy}{dx}+y=\tan^{-1}x

Write as:

\frac{dy}{dx}+\frac{1}{1+x^2}y=\frac{\tan^{-1}x}{1+x^2}

I.F.:

e^{\tan^{-1}x}

Answer:

ye^{\tan^{-1}x}=e^{\tan^{-1}x}(\tan^{-1}x-1)+C

Bernoulli DE Shortcut

Given:

\frac{dy}{dx}+P(x)y=Q(x)y^n

Use substitution:

z=y^{1-n}

Then solve linear equation in $z$ .

Exam case:

\frac{dy}{dx}+\frac{2y}{x}=\frac{y^3}{x^3}

Take:

z=y^{-2}

Answer:

x^{-4}y^{-2}=\frac{1}{3x^6}+C

Constant-Coefficient Linear DE

For

f(D)y = X

Steps:

Find C.F. from auxiliary equation.
Find P.I. from RHS.
Add: $y = \text{C.F.} + \text{P.I.}$
If IC given, use to find constants.

C.F. Rules

Distinct real roots $m_1,m_2$ : $y_c = C_1e^{m_1x}+C_2e^{m_2x}$
Repeated root $m$ : $y_c = (C_1+C_2x)e^{mx}$
Complex roots $\alpha \pm i\beta$ : $y_c = e^{\alpha x}(C_1\cos\beta x + C_2\sin\beta x)$

Hyper-Important Final Answers

With initial conditions

$y''+3y'+2y=0,\ y(0)=0,\ y'(0)=1$
$y=e^{-x}-e^{-2x}$
$(D^2-1)y=2,\ y(2)=-1,\ y'(2)=3$
$y=2e^{x-2}-e^{2-x}-2$
$(D^2-4D+4)y=x^2,\ y(0)=\frac38,\ y'(0)=1$
$y=\frac12 xe^{2x}+\frac14x^2+\frac12x+\frac38$
$y''+4y'+8y=e^{4x},\ y(0)=0,\ y'(0)=8$
$y=e^{-2x}\left(-\frac1{40}\cos 2x+\frac{157}{40}\sin 2x\right)+\frac{e^{4x}}{40}$

General solutions

$(D^2-3D+2)y=2\cosh x$
$y=C_1e^x+C_2e^{2x}-xe^x+\frac16e^{-x}$
$(D^2-4D+4)y=x^3$
$y=(C_1+C_2x)e^{2x}+\frac14x^3+\frac34x^2+\frac98x+\frac34$
$y''-3y'+4y=\cos 4x$
$y=e^{3x/2}\left(C_1\cos\frac{\sqrt7}{2}x+C_2\sin\frac{\sqrt7}{2}x\right)-\frac1{24}(\sin 4x+\cos 4x)$
$y''-4y'+4y=x^2+x$
$y=(C_1+C_2x)e^{2x}+\frac14x^2+\frac34x+\frac58$

Exam Hit List

ODE definition + applications
Order/degree
Forming DE by eliminating constants
Separation of variables using substitution
Exact equation test
Linear DE + integrating factor
Bernoulli reduction to linear form
C.F. and P.I. for constant-coefficient equations
Applying initial conditions fast

Applications of Differential Equations

Population growth/decay
Newton's law of cooling
Electrical circuits
Classical mechanics

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