Let y=y(x) be the solution of the differential equation [(dy)/(dx)+2y^2x = 2^2x+3 x^2x, y(…

Question

Let $y=y(x)$ be the solution of the differential equation
$$\frac{dy}{dx}+2y\sec^2x = 2\sec^2x+3\tan x\sec^2x,\qquad y(0)=\frac54.$$
Then $12\left(y\!\left(\dfrac{\pi}{4}\right)-e^{-2}\right)$ is equal to \_\_\_\_\_\_.

Answer: 21

Accepted Answer

I.F. $=e^{2\tan x}$. With $u=\tan x$:
$$ye^{2u} = \int e^{2u}(2+3u)\,du = e^{2u}+3\left(\frac{ue^{2u}}{2}-\frac{e^{2u}}{4}\right)+C.$$

Using $y(0)=5/4$: $C=1$. Then
$$y = 1+3\left(\frac{\tan x}{2}-\frac14\right)+e^{-2\tan x}.$$

At $x=\pi/4$: $y=\dfrac74+e^{-2}$. So
$$12\left(y\!\left(\frac{\pi}{4}\right)-e^{-2}\right) = 12\cdot\frac74 = 21.$$

Solutions · 1