IIT-JEE 1996Question 1
Determine the equation of the curve passing through the origin, in the form $y=f(x)$, which satisfies the differential equation $\dfrac{dy}{dx}=\sin(10x+6y)$.
Differential Equation
Answer$y=\dfrac13\left[\tan^{-1}\left(\dfrac{5\tan4x}{4-3\tan4x}\right)-5x\right]$
Solution
Concept. With $u=10x+6y$ the equation separates; $t=\tan\frac u2$ turns the integral into an inverse tangent. Formulas used. (1) $u=10x+6y\Rightarrow u'=10+6y'$(2) $\sin u=\dfrac{2t}{1+t^2}$, $du=\dfrac{2\,dt}{1+t^2}$, and $\int\dfrac{dt}{5t^2+6t+5}=\dfrac14\tan^{-1}\dfrac{5t+3}{4}$Step 1 - Separate. By (1), $u'=10+6\sin u$, so $\dfrac{du}{10+6\sin u}=dx$. Step 2 - Integrate. By (2), $\dfrac{du}{10+6\sin u}=\dfrac{dt}{5t^2+6t+5}$. At the origin $u=0$, $t=0$, so $$4x=\tan^{-1}\frac{5t+3}{4}-\tan^{-1}\frac34.$$ Step 3 - Solve for $t$. Take $\tan$ of both sides: $$\tan4x=\frac{\frac{5t+3}{4}-\frac34}{1+\frac{3(5t+3)}{16}}=\frac{4t}{5+3t}\ \Longrightarrow\ t=\frac{5\tan4x}{4-3\tan4x}.$$ Step 4 - Back to $y$. $u=2\tan^{-1}t$, so $10x+6y=2\tan^{-1}t$ and $$y=\frac13\left[\tan^{-1}\left(\frac{5\tan4x}{4-3\tan4x}\right)-5x\right].$$ Answer: $y=\dfrac13\left[\tan^{-1}\left(\dfrac{5\tan4x}{4-3\tan4x}\right)-5x\right]$