Question

The given family of functions is the general solution of the differential equation on the indicated interval.Find a member of the family that is a solution of the initial-value problem.

y=c1+c2cosx+c3sinx,(−[infinity],[infinity]) y=c 1 +c 2 cosx+c 3 sinx,(−[infinity],[infinity])
y'''+y′=0,y(π)=0,y′(π)=2,y''(π)=−1 y +y =0,y(π)=0,y ′ (π)=2,y ′′ (π)=−1

Answer 1

Step-by-step explanation:

`y'''+y=0---(i)`

General solution

`y=c_1e^o^x+c_2\cos x +c_3 \sin x\\\\\Rightarrow y=c_1+c_2 \cos x+c_3 \sin x---(ii)\\\\y(\pi)=0\\\\\Rightarrow 0=c_1+c_2\cos (\pi)+c_3\sin (\pi)\\\\\Rightarrow c_1-c_2=0\\\\c_1=c_2---(iii)`

`y'=-c_2\cos x+c_3\cosx\\\\y'(\pi)=2\\\\\Rightarrow2=-c_2\sin(\pi)+c_3\cos(\pi)\\\\\Rightarrow-c_2(0)+c_3(-1)=2\\\\\Rightarrow c_3=-2\\\\y''-c_2\cos x -c_3\sin x\\\\y''(\pi)=-1\\\\\Rightarrow-1=-c_2 \cos (\pi)=c_3\sin(\pi)\\\\\Rightarrow-1=c_2-0\\\\\Rightarrow c_2=-1`

in equation (iii)

`c_1=c_2=-1`

Therefore,

`\large\boxed{y=-1-\cos x-2\sin x}`