Browse Questions

Find the particular solution of the differential equation $(\tan^{-1}y-x)dy=(1+y^2)dx$,given that when x=0,y=0.

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Toolbox:
• A first order differential equation is an equation of the form $\frac{dy}{dx}+Px=Q$
• The solution is given by $ye^{\int Pdx}=\int Qe^{\int Pdx}dx+c$
• Note that $x$ and $y$ are interchangeable in the above equations.
• Integration by parts: $\int udv=uv-\int vdu$
• Integral of $e^x$ is $e^x$, i.e., $\int e^x dx = e^x + c$
• If $y=0, tan^{-1} y = 0$.
Given: $(\tan ^{-1}y -x) dy=(1+y^2)dx$
$\textbf{Step 1: Re-write the differential equation in the standard form}$:
$\Rightarrow$ $(\tan ^{-1}y -x) dy=(1+y^2)dx \rightarrow \large\frac{dx}{dy}=\frac{\tan^{-1}y}{1+y^2} - \frac{x}{1+y^2}$
$=>\large\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{\tan ^{-1}y}{1+y^2}$
This is a linear differential equation of the form: $\frac{dx}{dy}+Px=Q$.
The solution for this is given by $xe^{\int Pdy}=\int Qe^{\int Pdy}dy+c$
$\textbf{Step 2: Substitute for P and Q}$:
In the above equation, $P=\frac{1}{1+y^2}$ and $Q=\frac{\tan ^{-1}y}{1+y^2}$
$\Rightarrow P=\frac{1}{1+y^2}\rightarrow \int Pdy=\tan^{-1}y$