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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Solution of the differential equation $\large\frac{dy}{dx}+\frac{y}{x}$$=\sin x$ is:

$(A)\;x(y+\cos x)=\sin x+c \\ (B)\;x(y-\cos x)=\sin x+c \\ (C)\;xy\cos x=\sin x+c \\ (D)\;x(y+\cos x)=\cos x+c $

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  • A linear differential equation of the form $\large\frac{dy}{dx}$$+Py=Q$ has general solution as $ye^{\int pdx}=\int Qe^{\int pdx}dx+c$
  • $\int udv=uv-\int vdu$
Step 1:
Given $\large\frac{dy}{dx}+\frac{y}{x}$$=\sin x$
Clearly this is a linear differential equation of the form $\large\frac{dy}{dx}$$+Py=Q$. where $P=\large\frac{1}{x}$ and $Q= \sin x$
$\int pdx=\int \large\frac{1}{x} $$dx$
$=\log x$
Hence the integrating factor $e^{\large\int pdx}=e^{\large\log x}=x$ Hence the required solution is
$y \times I.F=\int Q.(I.F)dx+c$
$=y.x=\int x.\sin x dx+c$
Step 2:
Consider $\int x \sin x dx$
This is in the form $\int udx$
Let $ u=x,$ on differentiating w.r.t x
Let $dv=\sin x dx$, on integrating we get,
$v=-\cos x$
$\int udv=uv-\int vdu$
Now substituting $u,v,du$ and $dv$ we get
$\int x \sin x dx=x (-\cos x)-\int -\cos x.dx+c$
On integrating we get,
$-(x \cos x)+\sin x+c$
Hence the required solution is
$xy= -x \cos x +\sin x+c$
$=>x(y+\cos x)=\sin x+c$
Hence the correct option is $A$
answered May 21, 2013 by meena.p

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