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Exercise 9.5 · Q9

Q.Solve the following differential equation: xdydx+y−x+xycot⁡x=0(x≠0)x \frac{dy}{dx} + y - x + xy \cot x = 0 \quad (x \neq 0)

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Linear equation with integrating factor xsin⁡xx\sin x; the general solution is y=sin⁡x−xcos⁡x+Cxsin⁡xy=\dfrac{\sin x - x\cos x + C}{x\sin x}.

Spotting the type

yy and dydx\frac{dy}{dx} appear only to the first power, so this is first-order linear. We just need to get it into the standard shape dydx+P(x)y=Q(x)\frac{dy}{dx}+P(x)y=Q(x).

Standard form

Divide xdydx+y−x+xycot⁡x=0x\frac{dy}{dx} + y - x + xy\cot x = 0 by xx:

dydx+yx−1+ycot⁡x=0  ⟹  dydx+(1x+cot⁡x)y=1.\frac{dy}{dx} + \frac{y}{x} - 1 + y\cot x = 0 \implies \frac{dy}{dx} + \left(\frac{1}{x}+\cot x\right)y = 1.

So P=1x+cot⁡xP=\frac{1}{x}+\cot x and Q=1Q=1.

Integrating factor

∫(1x+cot⁡x)dx=log⁡∣x∣+log⁡∣sin⁡x∣=log⁡∣xsin⁡x∣,\int\left(\frac1x + \cot x\right)dx = \log|x| + \log|\sin x| = \log|x\sin x|,

so μ(x)=elog⁡∣xsin⁡x∣=xsin⁡x\mu(x)=e^{\log|x\sin x|} = x\sin x.

Multiply and integrate

The left side becomes an exact derivative:

ddx(y xsin⁡x)=xsin⁡x.\frac{d}{dx}\big(y\,x\sin x\big) = x\sin x. …

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