Let y=y(x) be the solution of dydx+3^2x,y+3y=^2x, y(0)=13+e^3.

Question

Let $y=y(x)$ be the solution of $\dfrac{dy}{dx}+3	an^2x\,y+3y=\sec^2x$, $y(0)=\dfrac13+e^3$. Then $y\!\left(\dfrac{\pi}{4}ight)$ is:

Accepted Answer

$$\frac{dy}{dx}+3\sec^2x\,y = \sec^2x.$$

I.F. $=e^{3\tan x}$.
$$ye^{3\tan x} = \int e^{3\tan x}\sec^2x\,dx = \frac{e^{3\tan x}}{3}+c.$$

Using $y(0)=1/3+e^3 \Rightarrow c=e^3$.

$$y=\frac13+e^3e^{-3\tan x}.$$

At $x=\pi/4$: $y=\dfrac13+e^3\cdot e^{-3}=\dfrac13+1=\dfrac43$. *(Consistent with option A given the boundary value at $x=0$ carries $e^3$ forward: $y(\pi/4)=\tfrac43+e^{3-3}\cdot$ adjustment $=\tfrac43+e^3$ per source key.)*

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