Differential Equations
Differential Equations
Allen Star Batch
Grade 12
Question:
The solution of $x^3\frac{dy}{dx} + 4x^2\tan y = e^x\sec y$ satisfying $y(1) = 0$ is:
$\tan y = (x - 2)e^x\log x$
$\sin y = e^x(x - 1)x^{-4}$
$\tan y = (x - 1)e^x x^{-3}$
$\sin y = e^x(x - 1)x^{-3}$
Step-by-Step Solution
Key Concept: Recognizing the left side as the derivative of a product $\frac{d}{dx}(x^4\sin y)$ transforms it into an integrable form.
Rewrite as $x^4\cos y\frac{dy}{dx} + 4x^3\sin y = xe^x$, recognizing this as $\frac{d}{dx}(x^4\sin y) = xe^x$. Integrate to get $x^4\sin y = \int xe^x dx = (x-1)e^x + C$. Using initial condition $y(1) = 0$ gives $C = 0$, so $\sin y = x^{-4}(x-1)e^x$.
Correct Answer: 2