Question

Let $y = y(x)$ be the solution curve of the differential equation $\sec y \frac{dy}{dx} + 2x \sin y = x^3 \cos y,$ $y(1) = 0$. Then $y\left(\sqrt{3}\right)$ is equal to:

• A. $\frac{\pi}{3}$
• B. $\frac{\pi}{6}$
• C. $\frac{\pi}{4}$
• D. $\frac{\pi}{12}$

Answer 1

Answer: (C)

$\frac{\pi}{4}$

Answer 2

Given the differential equation: $\sec^2 y \frac{dy}{dx} + 2x \sin y \sec y = x^3 \cos y.$

Rearranging terms: $\sec^2 y \frac{dy}{dx} + 2x \tan y = x^3.$ Let $t = \tan y$.

Then: $\sec^2 y \frac{dy}{dx} = \frac{dt}{dx}.$

Substituting into the equation: $\frac{dt}{dx} + 2xt = x^3.$

This is a linear first-order differential equation.

To solve it, we find the integrating factor (IF): $\text{IF} = e^{\int 2x dx} = e^{x^2}.$

Multiplying the entire equation by the integrating factor: $e^{x^2} \frac{dt}{dx} + 2x t e^{x^2} = x^3 e^{x^2}.$

This simplifies to: $\frac{d}{dx} (t e^{x^2}) = x^3 e^{x^2}.$

Integrating both sides: $t e^{x^2} = \int x^3 e^{x^2} \, dx.$

Using the substitution $z = x^2, \, dz = 2x dx$: $\int x^3 e^{x^2} \, dx = \frac{1}{2} \int z e^z \, dz.$

Integrating by parts: $\int z e^z \, dz = e^z (z - 1),$

so: $\frac{1}{2} \int z e^z \, dz = \frac{1}{2} e^{x^2} (x^2 - 1).$

Thus: $t e^{x^2} = \frac{1}{2} e^{x^2} (x^2 - 1) + C.$

Dividing by $e^{x^2}$: $t = \frac{1}{2} (x^2 - 1) + C e^{-x^2}.$

Recalling that $t = \tan y$,

we have: $\tan y = \frac{1}{2} (x^2 - 1) + C e^{-x^2}.$

Using the initial condition $y(1) = 0$: $\tan 0 = \frac{1}{2} (1^2 - 1) + C e^{-1^2},$ $0 = 0 + C e^{-1} \implies C = 0.$

Thus: $\tan y = \frac{1}{2} (x^2 - 1).$

To find $y(\sqrt{3})$: $\tan y = \frac{1}{2} ((\sqrt{3})^2 - 1) = \frac{1}{2} (3 - 1) = 1.$

Therefore: $y = \tan^{-1} (1) = \frac{\pi}{4}.$

Therefore: $\frac{\pi}{4}.$