Let g be a differentiable function such that _0^x g(t),dt = x-_0^x t,g(t),dt, x0, and let …

Question

Let $g$ be a differentiable function such that $\int_0^x g(t)\,dt = x-\int_0^x t\,g(t)\,dt$, $x\ge0$, and let $y=y(x)$ satisfy $\dfrac{dy}{dx}-y	an x=2(x+1)\sec x\,g(x)$, $x\in[0,\pi/2)$, $y(0)=0$. Then $y\!\left(\dfrac{\pi}{3}ight)$ is:

Accepted Answer

Differentiating the integral equation: $g(x)=1-xg(x) \Rightarrow g(x)=\dfrac{1}{1+x}$.

So the ODE becomes $\dfrac{dy}{dx}-y\tan x = 2\sec x$.

I.F. $=\cos x$. $\dfrac{d}{dx}(y\cos x)=2 \Rightarrow y\cos x=2x+c$. Using $y(0)=0 \Rightarrow c=0$.

$$y=\frac{2x}{\cos x} \implies y\!\left(\frac{\pi}{3}\right) = \frac{2\pi/3}{1/2} = \frac{4\pi}{3}.$$

Solutions · 1