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Question

# Find the general solution of the differential equation xdydx+y−x+xy cot x=0 (x≠0)

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Solution

## Given differential equation : xdydx+y−x+xy cot x=0 dydx+yx−1+y cot x=0 dydx+y(1x+cot x)−1=0 dydx+(1x+cot x)y=1 Given differential equation is of the form dydx+Py=Q By comparing both the equations, we get P=(1x+cot x) and Q=1 The general solution of the given differential equation is y(I.F)=∫(Q×I.F.)dx+c ....(i) Firstly, we need to find I.F. I.F.=e∫pdx I.F.=e∫⎛⎝1x+cot x⎞⎠dx I.F.=elog x+log sin x I.F.=elog(x sin x) (∵log a+log b=log ab) I.F.=x sin x (∵elog x=x) Substituting the value of I.F in (i), we get y.x sin x=∫1.x sin x dx+c Using Integration by parts : ∫f(x)g(x)dx=f(x)∫g(x)dx−f[f′(x)∫g(x)dx]dx Taking f(x)=x and g(x)=sin x y x sin x=x∫sin x dx−∫[1.∫sin x.dx]dx+c y x sin x=x(−cos x)−∫1.(−cos x)dx+c y x sin x=−x cos x+∫cos x dx+c y x sin x=−x cos x+sin x+c y=−x cos xx sin x+sin xx sin x+cx sin x y=−cot x+1x+cx sin x Hence, the required general solution is y=−cot x+1x+cx sin x ∴y=(1+x2)−1log |sin x|+c(1+x2)−1

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