Differential Equations
Bernoulli DE — Substitution v = tan y
nta_pyq_2026_jan
Grade None
Question:
Let $y=y(x)$ be the solution of the differential equation $x\dfrac{dy}{dx}-\sin 2y=x^3\left(2-x^3\right)\cos^2 y$, $x\neq0$. If $y(2)=0$, then $\tan(y(1))$ is equal to
-\dfrac{3}{4}
\dfrac{3}{4}
-\dfrac{7}{4}
\dfrac{7}{4}
Step-by-Step Solution
Key Concept: Divide by $\cos^2 y$: $x\sec^2 y\,\dfrac{dy}{dx}-2\tan y=x^3(2-x^3)$. Let $v=\tan y$: $x\dfrac{dv}{dx}-2v=x^3(2-x^3)$. Divide by $x$: $\dfrac{dv}{dx}-\dfrac{2v}{x}=x^2(2-x^3)$. IF $=x^{-2}$.
$\tan y=2x^3-\tfrac{x^6}{4}$. $\tan(y(1))=\tfrac{7}{4}$.
Correct Answer: 4