Solving Differential Equations
Tips
- \(\frac{dy}{dx}=f(x)g(y)\)
- \(\int \frac{1}{g(y)} \, dy=\int f(x) \, dx\)
- Need to add \(+c\) to both sides of the equation after integration
- The family of solutions of a differential equation is the graphical representation of the curve with different values of \(c\)
e.g. \(\frac{dy}{dx}=12x^2-1\)
\(\int 1 \, dy=\int (12x^2-1) \, dx\)
\(y=4x^3-x+c\)
\(=x(2x-1)(2x+1)+c \to\) This is a general solution as we do not know what \(c\) is.
- If we know a point on the curve of the particular solution we want to find, you can find what \(c\) is.
- This point is called a boundary condition
Find the general solution to \((1+x^2)\frac{dy}{dx}=x\tan y\)
- \(\frac{1}{\tan y}dy=\frac{x}{1+x^2}dx\)
- \(\int \frac{1}{\tan y} \, dy=\int \frac{x}{1+x^2} \, dx\)
- \(\int \cot y \, dy=\int \frac{x}{1+x^2} \, dx\)
- \(\ln |\sin y|=\frac{1}{2}\ln |1+x^2|+\ln k\)
- \(\ln |\sin y|=\ln k\sqrt{ 1+x^2 }\)
- \(\sin y=k\sqrt{ 1+x^2 }\)
- \(y=\arcsin(k\sqrt{ 1+x^2 })\)
Find the particular solution to \(\frac{dy}{dx}=-\frac{3(y-2)}{(2x+1)(x+2)}\) given that \(x=1\) when \(y=4\)
- \(\frac{1}{y-2}dy=-\frac{3}{(2x+1)(x+2)}dx\)
- \(-\frac{3}{(2x+1)(x+2)}=\frac{A}{2x+1}+\frac{B}{x+2}\)
- \(-3=A(x+2)+B(2x+1)\)
- \(x=-2\)
- \(-3=-3B\to B=1\)
- \(x=-\frac{1}{2}\)
- \(-3=\frac{3}{2}A\to A=-2\)
- \(\int \frac{1}{y-2} \, dy=\int \left( -\frac{2}{2x+1}+\frac{1}{x+2} \right) \, dx\)
- \(\ln |y-2|=-\ln |2x+1|+\ln |x+2|+\ln k\)
- Since \(x=1\), \(y=4\)
- \(\ln2=-\ln3+\ln3+\ln k\)
- \(\ln2=\ln k\)
- \(k=2\)
- \(\ln |y-2|=\ln |\frac{k(x-2)}{2x+1}|\)
- \(y-2=\frac{2(x+2)}{2x+1}\)
- \(y=\frac{2x+4}{2x+1}+2\)
- Top heavy fraction so simplify:
- \(y=\frac{3}{2x+1}+3\)
Find the general solution to \(\frac{dy}{dx}=(1+y)(1-2x)\)
- \(\int \frac{1}{1+y} \, dy=\int 1-2x \, dx\)
- \(\ln |1+y|=x-x^2+c\)
- \(1+y=e^{x-x^2+c}\)
- \(y=e^{x-x^2+c}-1\)
- \(B=e^c\)
- \(y=Be^{x-x^2}-1\)
Find the general solution to \(\frac{dy}{dx}=y\tan x\)
- \(\int \frac{1}{y} \, dy=\int \tan x \, dx\)
- \(\dots\)