Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDifferential Equations3 Marks
Question
Solve the differential equation $\left(\tan ^{-1} y-x\right) d y=\left(1+y^2\right) d x$
✓
Answer
Rewrite as a linear differential equation in $x: \frac{d x}{d y}+\frac{x}{1+y^2}=\frac{\tan ^{-1} y}{1+y^2}$ This is of the form $\frac{d x}{d y}+P x=Q$, where $P=\frac{1}{1+y^2}$ and $Q=\frac{\tan ^{-1} y}{1+y^2}$. Integrating Factor (I.F.): $e^{\int P d y}=e^{\int \frac{1}{1+y^2} d y}=e^{\tan ^{-1} y}$ $x(I . F)=.\int(Q$.I.F.) $d y+C x e^{\tan ^{-1} y}$ $=\int \frac{\tan ^{-1} y}{1+y^2} e^{\tan ^{-1}} y d y$ Let $\tan ^{-1}$ $y=t \Longrightarrow \frac{1}{1+y^2} d y=d t x e^t=$ $\int t e^t d t=(t-1) e^t+C$ $x=\tan ^{-1} y-1+C e^{-\tan ^{-1} y}$
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