Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSModel Paper 53 Marks
Question
Solve: $\frac{d y}{d x}+\frac{y}{x}=\cos x +\frac{\sin x}{x}$
✓
Answer
We have,
$\frac{d y}{d x}+\left(\frac{1}{x}\right) y=\cos x+\frac{\sin x}{x} \ldots$
This is a linear differential equation of the form
$\frac{d y}{d x}+ Py = Q$, where $P =\frac{1}{x}$ and $Q =\cos x +\frac{\sin x}{x}$
$\therefore$ I. $F=e^{\int P d x}=e^{\int \frac{1}{x} d x}=e^{\log x}=x$
Multiplying both sides of $(i)$ by $I.F. = x,$ we get
$x \frac{d y}{d x}+y=x \cos x+\sin x$
Integrating both sides with respect to $x,$ we get
$y x=\int(x \cos x+\sin x) d x+C\left[\right.$ Using: $y=( I.F. )=\int Q( I.F. \left.) d x+C\right]$
$\Rightarrow xy =\int \underset{I}{x} \cos xdx +\int \sin x dx + C$
$\Rightarrow x y=x \sin x-\int \sin x d x+\int \sin x d x+C\ [$Integrating $1^{st}$ integral by parts$]$
$\Rightarrow x y=x \sin x+C$
$\Rightarrow y=\sin x+\frac{C}{x}$
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