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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