Differential Equations

# Linear Form

Linear form is a differential equation of the first order and first degree in the form $\frac{dy}{dx}+Py=Q$ where $P$ and $Q$ are the functions of $x$ or constants (but not of $y$). For example, the equation $\frac{dy}{dx}+xy=x^2$ is linear. The linear equation $\frac{dy}{dx}+Py=Q$ becomes easily integrable if its both sides are multiplied by $e^{\int Pdx}$. This quantity or factor is called the Integration Factor (I.F.) of the linear equation.

Sometimes, we may come across a differential equation of the form $\frac{dy}{dx}+Py=Qy^2$

This equation can be changed into the form $y^{-2}\frac{dy}{dx}+Py^{-1}=Q$

Put $y^{-1}=z$ so that $y^{-2}\frac{dy}{dz}=-\frac{dz}{dx}$ $\therefore\frac{dz}{dx}-Pz=-Q$ which is the differential equation of the linear form.

## Solve the following equations:

### $\frac{dy}{dx}+\frac{1}{x}y=x^2$

$\frac{dy}{dx}+\frac{1}{x}y=x^2\text{ __(1)}$ Comparing this equation with $\frac{dy}{dx}+Py=Q$, we get, $P=\frac{1}{x} \text{ and }Q=2x^2$ $\therefore I.F.=e^{\int Pdx}=e^{\int\frac{1}{x}dx}=e^{\log x}=x$

Multiplying $\text{(1)}$ by $I.F.$, we get, $x\frac{dy}{dx}+y=x^3$ $xdy+ydx=x^3dx$ $d(xy)=x^3dx$ Integrating both sides, we get, $xy=\frac{1}{4}x^4+C$ where $C$ is an arbitrary constant.

### $x\frac{dy}{dx}-y=2x^3$

$x\frac{dy}{dx}-y=2x^3$ $\frac{dy}{dx}-\frac{y}{x}=2x^2\text{ __(1)}$ Comparing this equation with $\frac{dy}{dx}+Py=Q$, we get $P=-\frac{1}{x}\text{ and }Q=2x^2$ $\therefore I.F.=e^{\int Pdx}=e^{-\int\frac{1}{x}dx}=e^{-\log x}=\frac{1}{x}$

Multiplying $\text{(1)}$ by $I.F.$, we get $\frac{1}{x}\frac{dy}{dx}-\frac{y}{x^2}=2x$ $d\left(y.\frac{1}{x}\right)=2xdx$ Integrating both sides, we get, $\frac{y}{x}=\int 2xdx$ $\frac{y}{x}=x^2+C$ $\therefore y=x^3+Cx$

### $(x+1)\frac{dy}{dx}+2y=\frac{e^x}{(x+1)}$

$(x+1)\frac{dy}{dx}+2y=\frac{e^x}{(x+1)}$ $\frac{dy}{dx}+\frac{2y}{(x+1)}=\frac{e^x}{(x+1)^2}\text{ __(1)}$ Comparing this equation with $\frac{dy}{dx}+Py=Q$, we get $\therefore P=\frac{2}{(x+1)}\text{ and }Q=\frac{e^x}{(x+1)^2}$ $\therefore I.F.=e^{\int Pdx}=e^{\int\frac{2}{(x+1)}dx}=e^{2\log(x+1)}=(x+1)^2$

Multiplying $\text{(1)}$ by $I.F.$, we get, $(x+1)^2\frac{dy}{dx}+(x+1)2y=e^x$ $d[(x+1)^2y]=e^xdx$ Integrating both sides, we get $(x+1)^2=\int e^xdx$ $\therefore(x+1)^2y=e^x+C$

### $\sin x\frac{dy}{dx}+y\cos x=\sin x\cos x$

$\sin x\frac{dy}{dx}+y\cos x=\sin x\cos x$ $\frac{dy}{dx}+\frac{\cos x}{\sin x}y=\cos x\text{ __(1)}$ Comparing this equation with $\frac{dy}{dx}+Py=Q$, we get $\therefore P=\frac{\cos x}{\sin x}\text{ and }Q=\cos x$ $\therefore I.F.=e^{\int Pdx}=e^{\int\frac{\cos x}{\sin x}dx}=e^{\log\sin x}=\sin x$

Multiplying $\text{(1)}$ by $I.F.$, we get $\sin x\frac{dy}{dx}+y\cos x=\sin x\cos x$ $d(\sin x.y)=\sin x\cos xdx$ Integrating both sides, we get $y\sin x=\int\sin x\cos xdx$ $y\sin x=\frac{1}{2}\int\sin 2xdx$ $y\sin x=-\frac{1}{4}\cos 2x+C$ $\therefore y\sin x+\frac{1}{4}\cos 2x=C$

### $(1-x^2)\frac{dy}{dx}-xy=1$

$(1-x^2)\frac{dy}{dx}-xy=1$ $\frac{dy}{dx}-\frac{x}{(1-x^2)}y=\frac{1}{(1-x^2)}\text{ __(1)}$ Comparing this equation with $\frac{dy}{dx}+Py=Q$, we get $P=\frac{-x}{(1-x^2)}\text{ and }Q=\frac{1}{(1-x^2)}$ $\therefore I.F.=e^{\int Pdx}=e^{\int\frac{-x}{1-x^2}}dx=e^{\frac{1}{2}\int\frac{-2x}{1-x^2}dx}$$=e^{\frac{1}{2}\log(1-x^2)}=\sqrt{1-x^2}$

Multiplying $\text{(1)}$ by $I.F.$, we get $\sqrt{1-x^2}\frac{dy}{dx}-\frac{x}{\sqrt{1-x^2}}y=\frac{1}{\sqrt{1-x^2}}$ $d\left(y\sqrt{1-x^2}\right)=\frac{dx}{\sqrt{1-x^2}}$ Integrating both sides, we get, $y\sqrt{1-x^2}=\int\frac{dx}{\sqrt{1-x^2}}$ $\therefore y\sqrt{1-x^2}=\sin^{-1}x+C$

### $\frac{dy}{dx}+\frac{2x}{1+x^2}y=\frac{1}{(1+x^2)^2}$

$\frac{dy}{dx}+\frac{2x}{1+x^2}y=\frac{1}{(1+x^2)^2}\text{ __(1)}$ Comparing this equation with $\frac{dy}{dx}+Py=Q$, we get $P=\frac{2x}{1+x^2}\text{ and }Q=\frac{1}{(1+x^2)^2}$ $\therefore I.F.=e^{\int Pdx}=e^{\int\frac{2x}{1+x^2}dx}=e^{\log(1+x^2)}=1+x^2$

Multiplying $\text{(1)}$ by $I.F.$, we get, $(1+x^2)\frac{dy}{dx}+2xy=\frac{1}{(1+x^2)}$ $d[(1+x^2)y]=\frac{dx}{1+x^2}$ Integrating both sides, we get $(1+x^2)y=\int\frac{dx}{1+x^2}$ $\therefore(1+x^2)y=\tan^{-1}x+C$

### $x\frac{dy}{dx}+2y=x^2\log x$

$x\frac{dy}{dx}+2y=x^2\log x$ $\frac{dy}{dx}+\frac{2}{x}y=x\log x\text{ __(1)}$ Comparing this equation with $\frac{dy}{dx}+Py=Q$, we get $\therefore P=\frac{2}{x}\text{ and }Q=x\log x$ $\therefore I.F.=e^{\int Pdx}=e^{\int\frac{2}{x}dx}=e^{2\log x}=x^2$ Multiplying $\text{(1)}$ by $I.F.$, we get $x^2\frac{dy}{dx}+2xy=x^3\log x$ $d(x^2y)=x^3\log xdx$ Integrating both sides, we get $x^2y=\int x^3\log xdx$ $x^2y=\log x\int x^3dx-\int\left(\frac{d}{dx}(\log x)\int x^3dx\right)dx$ $x^2y=\log x\frac{x^4}{4}-\int\left(\frac{1}{x}.\frac{x^4}{4}\right)dx$ $x^2y=\frac{1}{4}x^4\log x-\frac{1}{16}x^4+C$ $\therefore y=\frac{1}{4}x^4\log x-\frac{1}{16}x^2+\frac{C}{x^2}$

Also see: Integration by Parts

### $\frac{dy}{dx}+y=xy^2$

$\frac{dy}{dx}+y=xy^2$ $\frac{1}{y^2}\frac{dy}{dx}+\frac{1}{y}=x\text{ __(1)}$ Put $\frac{1}{y}=z$ so that $-\frac{1}{y^2}\frac{dy}{dx}=\frac{dz}{dx}$ $\Rightarrow \frac{1}{y^2}\frac{dy}{dx}=-\frac{dz}{dx}$ Then, equation $\text{(1)}$ becomes $-\frac{dx}{dx}+z=x$ $\frac{dz}{dx}-z=-x\text{ __(2)}$ Comparing this equation with $\frac{dz}{dx}+Py=Q$, we get $\therefore P=-1\text{ and }Q=-x$ $\therefore I.F.=e^{\int Pdx}=e^{\int -1dx}=e^{-x}$

Multiplying $\text{(2)}$ by $I.F.$, we get $e^{-x}\frac{dz}{dx}-e^{-x}z=-xe^{-x}$ $d(e^{-x}z)=-xe^{-x}dx$ Integrating both sides, we get, $e^{-x}z=\int -xe^{-x}dx$ $e^{-x}z=-\left[x\int e^{-x}dx-\int\left(\frac{d}{dx}(x)\int e^{-x}dx\right)dx\right]$ $e^{-x}\frac{1}{y}=-\left[-xe^{-x}-e^{-x}\right]+C$ $\frac{e^{-x}}{y}=xe^{-x}+e^{-x}+C$ $\therefore xy+y+Cye^x=1$

### $\frac{dy}{dx}+y\tan x=y^3\sec x$

$\frac{dy}{dx}+y\tan x=y^3\sec x$ $\frac{1}{y^3}\frac{dy}{dx}+\frac{1}{y^2}\tan x=\sec x\text{ __(1)}$ Put $\frac{1}{y^2}=z$, then $-\frac{2}{y^3}\frac{dy}{dx}=\frac{dz}{dx}$ $\Rightarrow\frac{1}{y^3}\frac{dy}{dx}=-\frac{1}{2}\frac{dz}{dx}$ Then, equation $\text{(1)}$ becomes $-\frac{1}{2}\frac{dz}{dx}+z\tan x=\sec x$ $\frac{dz}{dx}-2z\tan x=-2\sec x\text{ __(2)}$ Comparing this equation with $\frac{dy}{dx}+Py=Q$, we get $\therefore P=-2\tan x\text{ and }Q=-2\sec x$ $\therefore I.F.=e^{\int Pdx}=e^{\int -2\tan xdx}=e^{2\log\cos x}=cos^2x$

Multiplying $\text{(2)}$ by $I.F.$, we get $\cos^2x\frac{dz}{dx}-2z\cos^2x\tan x=-2\sec x\cos^2x$ $\cos^2x\frac{dz}{dx}-2z\cos x\sin x=-2\cos x$ $d(z\cos^2x)=-2\cos xdx$ Integrating both sides, we get $z\cos^2x=-2\int\cos xdx$ $z\cos^2x=-2\sin x+C$ $\frac{1}{y^2}\cos^2x=-2\sin x+C$ $\therefore\cos^2x=(C-2\sin x)y^2$

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