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Find the general solution of the differential equation $x\large\frac{dy}{dx}$$+y-x+xy\cot x=0\;(x\neq0) Can you answer this question? 1 Answer 0 votes Toolbox: • The first order linear equation can be solved as: • (i) By writing the equation as \large\frac{dy}{dx}$$ +Py = Q$
• (ii) Find the integral factor(I.F) $\int Pdx = e^{\int Pdx}$
• (iii) Write the solution as $y. (I.F) = \int Q(I.F)dx + C$
Step 1:
Using the information in the tool box, let us write the equation as $\large\frac{dy}{dx} =\frac{ (1+x\cot x)y}{x} $$= 1 Here P = (\large\frac{1}{x})$$ + \cot x$ and $Q = 1$
$\int pdx = \int [\large\frac{1}{x}$$+ \cot x]dx \qquad\;\;= \log x + \log|\sin x| I.F = e^[\log\sin x + \log x] = e^{\log[x\sin x] }= [x\sin x] The required solution is y.\log[x\sin x] = \int1.[x\sin x] + C Step 2: Integration of [x\sin x] can be done by parts. Let u = x then du = dx and dv = \sin xdx, v = -\cos x \int x\sin x = x(-\cos x) - \int(-\cos x).dx \qquad\;\;\;\;\;= -[x\cos x] +\int\cos xdx \qquad\;\;\;\;\;= -[x\cos x] +\sin x + C Dividing throughout by x\sin x we get y = -\cot x + \large\frac{1}{x}$$+ C$
This is the required solution.