InterviewSolution
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InterviewSolution
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Solve the following differential equation:(y - sin2x)dx + tanx dy = 0 |
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Answer» (y - sin2x)dx + tanx dy = 0 \(\frac{dy}{dx} = \frac{sin^2x - y}{tanx}\) ⇒ \(\frac{dy}{dx} + \frac y{tanx} = \frac{sin^2x}{tanx} = sinx. cos x\) which is linear differential equation \(\therefore I.F = e^{\int P dx}\) \(= e^{\int\frac1{tanx}dx}\) \(= e^{\int cot x\,dx}\) \(= e^{log\, sinx}\) \(= sin x\) \(\therefore y \times(I.F) = \int Q \times (I.F)dx\) \(y \times sin x= \int (sinx \,cos x)sinx\,dx\) \(= \int sin^2x \,cosx\,dx\) \(= \frac{sin^3x}{3} + C\) ⇒ \(y = \frac{sin^2x}{3} + \frac C{sin\,x}\) ⇒ \(y = \frac{sin^2x}{3 } + C \,cosecx\) is solution of given differential equation. |
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