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Solve:$y+\large\frac{d}{dx}$$(xy)=x(\sin x+log x)$

$\begin{array}{1 1}(A)\;\cos x -\large\frac{2 \sin x}{x}+\frac{2 \cos x}{x^2}+\frac{x \log x }{3}-\frac{x}{9}+\frac{c}{x^2} \\(B)\;-\cos x +\large\frac{2 \sin x}{x}+\frac{2 \cos x}{x^2}+\frac{x \log x }{3}-\frac{x}{9}+\frac{c}{x^2} \\(C)\;\cos x +\large\frac{2 \sin x}{x}-\frac{2 \cos x}{x^2}-\frac{x \log x }{3}-\frac{x}{9}+\frac{c}{x^2} \\ (D)\;=-\cos x +\large\frac{2 \sin x}{x}+\frac{2 \cos x}{x^2}-\frac{x \log x }{3}-\frac{x}{9}-\frac{c}{x^2} \end{array} $

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  • A linear differential equation of the form $\large\frac{dy}{dx}$$+Py=Q$ has the general equation as $ye^{\int pdx}=\int Q .e^{\int pdx} dx+c$ where $e^{\int pdx}$ is the integrating factor (I.F)
Step 1:
Given $y+\large\frac{d}{dx}$$(xy)=x(\sin x+\log x)$
$\large\frac{d}{dx}$$(xy)$ is in the product form
which$\quad= \large\frac{d}{dx}$$(x).y+\large\frac{d}{dx}(y).x$
$\qquad\quad=y+x \large\frac{dy}{dx}$
Hence the equation can be written as
$y+y+x \large\frac{dy}{dx}$$=x(\sin x+\log x)$
$2y+x \large\frac{dy}{dx}$$=x(\sin x+\log x)$
Dividing throughout by x we get
$\large\frac{dy}{dx}+\frac{2}{x}$$.y=\sin x+\log x$
Clearly this represents a linear differential equation of the form
$\large\frac{dy}{dx}$$+Py=Q$ where $P=\large\frac{2}{x}$ and $Q=\sin x+\log x$
Next let us find the integrating factor I.F which is $e^{\int pdx}$
$\int pdx=\int \large\frac{2}{x} $$dx$
$2 \int \large\frac{1}{x}$$dx= 2\log x$
$=\log x^2$
Hence $ e^{\int pdx}=e^{\log x^2}$
But $e^{\log x}=x$
Similarly $e^{\log x^2}=x^2$
Hence $I.F=x^2$
Now the required solution is
$y \times I.F=\int Q \times I.F\; dx+c$
$=>y \times x^2=\int (\sin x+\log x).x^2dx+c$
$=>y x^2=\int x^2 \sin x dx +\int x^2 \log xdx+c$
Step 2:
Consider $\int x^2. \sin x dx$
This is of the form $\int udv=uv-\int vdu$
Let $ u=x^2$ on differentiating we get, $du=2xdx$
Let $ dv=\sin xdx $ on integrating we get, $ v=-\cos x$
Now substituting for $u,v,du$ and $dv$ we get
$x^2(-\cos x)-\int -\cos x.2xdx$
$=-x^2\cos x+2 \int x \cos x dx$
Again consider $\int x \cos x dx$
Let $ u=x$ on differentiating w.r.t x we get, $du=dx$
Let $ dv=\cos xdx $ on integrating we get, $ v=\sin x$
Therefore $\int x \cos x dx= x\sin x-\int \sin x dx$
On integrating we get,
$x \sin x-(-\cos x)$
$=x\sin x+\cos x$
Hence $ \int x^2 \sin x dx =-x^2\cos x+2 (x \sin x+\cos x)$
$=-x^2\cos x+2x \sin x +2 \cos x$-----(1)
Step 3:
Next consider $\int x^2. \log x dx$
This is also of the form $\int udv$
Similarly let us take
$dv=x^2$, on integrating we get,
Let $u=\log x$
On differentiating we get, $du=\large\frac{1}{x}$$dx$
Hence substituting $u,v,du$ and $dv$ as before
$\large\frac{x^3\log x}{3}-\frac{1}{3} \int $$x^3.\large \frac{1}{x}$$dx$
$\large\frac{x^3\log x}{3}-\frac{1}{3} \int $$x^2 \;dx$
On integrating we get,
$\large\frac{x^3 \log x}{3}-\frac{x^3}{9}$$+c$-----(2)
Now combining equ(1) and equ(2) in the required solution we get,
$y.x^2=-x^2\cos x+2 x \sin x +2 \cos x +\large\frac{x^3 \log x}{3}-\frac{x^3}{9}$$+c$
Dividing throughout by $x^2$ we get
$y=-\cos x +\Large\frac{2 \sin x}{x}+\frac{2 \cos x}{x^2}+\frac{x \log x }{3}-\frac{x}{9}+\frac{c}{x^2}$
is the required solution


answered May 24, 2013 by meena.p
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