Question

The general solution of $\left(x \frac{dy}{dx} - y\right) \sin \frac{y}{x} = x^3 e^x$ is

• A. $e^x(x - 1) + \cos \frac{y}{x} + c = 0$
• B. $x e^x + \cos \frac{y}{x} + c = 0$
• C. $e^x(x + 1) + \cos \frac{y}{x} + c = 0$
• D. $e^x x - \cos \frac{y}{x} + c = 0$

Hint:

Whenever you see the expression $(x dy - y dx)$ or $(x y' - y)$, immediately think of dividing by $x^2$ to create the exact derivative $d(y/x)$, or dividing by $y^2$ to create $-d(x/y)$. It is one of the most powerful shortcuts in differential equations.

Answer 1

The Correct Option is A

Step 1: Look at the equation.
We have $\left(x\frac{dy}{dx}-y\right)\sin\frac{y}{x}=x^3 e^x$. The piece $x\frac{dy}{dx}-y$ and the term $\frac{y}{x}$ are a strong hint that $\frac{y}{x}$ should be treated as one quantity.

Step 2: Recall a handy derivative.
By the quotient rule, \[ \frac{d}{dx}\!\left(\frac{y}{x}\right)=\frac{x\frac{dy}{dx}-y}{x^2}. \] So $x\frac{dy}{dx}-y$ is just $x^2$ times the derivative of $\frac{y}{x}$.

Step 3: Divide to reveal that pattern.
Divide both sides of the equation by $x^2$: \[ \frac{x\frac{dy}{dx}-y}{x^2}\sin\frac{y}{x}=x e^x. \] The left side now starts with $\frac{d}{dx}\!\left(\frac{y}{x}\right)$.

Step 4: Substitute one new letter.
Let $v=\frac{y}{x}$. Then the equation becomes \[ \sin v\,\frac{dv}{dx}=x e^x, \] so $\sin v\,dv=x e^x\,dx$. The variables are now separated.

Step 5: Integrate both sides.
Left side: $\int\sin v\,dv=-\cos v$. Right side: $\int x e^x\,dx$ needs integration by parts with $u=x$, $dv=e^x dx$, giving $x e^x-\int e^x dx=x e^x-e^x$.

Step 6: Combine the results.
\[ -\cos v=x e^x-e^x+C_1=e^x(x-1)+C_1. \]

Step 7: Put $v=\frac{y}{x}$ back and tidy up.
Move everything to one side: \[ e^x(x-1)+\cos\frac{y}{x}+c=0. \] This is option (1).
\[ \boxed{e^x(x-1)+\cos\dfrac{y}{x}+c=0} \]