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

Answers (1)

y+\frac{dy}{dx}(xy)=x(\sin x +\log x)

Now dx/dy (xy) refers to the differentiation of xy with respect to x

Using product rule

$$ \Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{xy})=\mathrm{y}+\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}} $$
When we put it back originally in the differential equation given,

$$\\ \Rightarrow y+y+x \frac{d y}{d x}=x(\sin x+\log x) \\ \Rightarrow x \frac{d y}{d x}+2 y=x(\sin x+\log x) $$
Divide by x
$$ \Rightarrow \frac{d y}{d x}+\left(\frac{2}{x}\right) y=\sin x+\log x $$
Compare \frac{\mathrm{dy}}{\mathrm{dx}}+\left(\frac{2}{\mathrm{x}}\right) \mathrm{y}=\sin \mathrm{x}+\log _{\mathrm{x}}$ with $ \quad \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q}$
We get
$$ \mathrm{P}=\frac{2}{\mathrm{x}} \text { and } \mathrm{Q}=\sin \mathrm{x}+\log \mathrm{x} $$

The above equation is a linear differential equation with P and Q as    functions of x

The first to find the solution of a linear differential equation is to find the integrating factor. \Rightarrow \mathrm{IF}=\mathrm{e}^{[\mathrm{Pdx}}$
$$ \\ \Rightarrow \mathrm{IF}=\mathrm{e}^{2 \int \frac{1}{\mathrm{x}} \mathrm{dx}} \\ \Rightarrow \mathrm{IF}=\mathrm{e}^{2 \log \mathrm{x}} \\ \Rightarrow \mathrm{IF}=\mathrm{e}^{\log \mathrm{x}^{2}}=\mathrm{x}^{2} \\ \Rightarrow \mathrm{IF}=\mathrm{x}^{2} $$
The solution of the linear differential equation is
y(\mathrm{IF})=\int Q(\mathrm{IF}) \mathrm{d} \mathrm{x}+\mathrm{c}$
Substituting values for Q and IF
\Rightarrow y x^{2}=\int(\sin x+\log x) x^{2} d x+c$
$$ \Rightarrow y x^{2}=\int x^{2} \sin x d x+\int x^{2} \log x d x+c \ldots(a) $$

Find the integrals individually,

Using uv for integration

$$ \\ \Rightarrow \int_{U V} d x=u \int v d x-\int\left(u^{\prime} j v\right) d x \\ \Rightarrow \int x^{2} \sin x d x=x^{2}(-\cos x)-\int 2 x(-\cos x) d x $$
\Rightarrow \int x^{2} \sin x d x=-x^{2} \cos x+2 \int x \cos x d x$
$$ \Rightarrow \int x^{2} \sin x d x=-x^{2} \cos x+2\left(x \sin x-\int \sin x d x\right) $$
\Rightarrow \int x^{2} \sin x d x=-x^{2} \cos x+2(x \sin x-(-\cos x))$
$$ \Rightarrow \int x^{2} \sin x d x=-x^{2} \cos x+2 x \sin x+2 \cos x \ldots \text { (i) } $$
Now
Use product rule
$$ \\ \Rightarrow \int x^{2} \log x d x=\log x\left(\frac{x^{3}}{3}\right)-\int\left(\frac{1}{x}\right)\left(\frac{x^{3}}{3}\right) d x \\ \Rightarrow \int x^{2} \log x d x=\left(\frac{x^{3}}{3}\right) \log x-\frac{1}{3} \int x^{2} d x $$

$$ \Rightarrow \int x^{2} \log x d x=\left(\frac{x^{3}}{3}\right) \log x-\frac{x^{3}}{9} \ldots(i i) $$
Substitute (i) and (ii) in (a)
$$ \Rightarrow \mathrm{yx}^{2}=-\mathrm{x}^{2} \cos \mathrm{x}+2 \mathrm{x} \sin \mathrm{x}+2 \cos \mathrm{x}+\left(\frac{\mathrm{x}^{3}}{3}\right) \log \mathrm{x}-\frac{\mathrm{x}^{3}}{9}+\mathrm{c} $$
Divide by  x^{2}$
$$ \Rightarrow y=-\cos x+\frac{2 \sin x}{x}+\frac{2 \cos x}{x^{2}}+\left(\frac{x}{3}\right) \log x-\frac{x}{9}+\frac{c}{x^{2}} $$

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