What is the solution for y^{''}+y=2,+ty(0)=1,y(0)=1 ?

The solution for y^{''}+y=2,+ty(0)=1,y(0)=1 is y=c_{1}cos(t)+c_{2}sin(t)+2

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y^{''}+y=2,+ty(0)=1,y(0)=1

y^{^{''}} + y = 2, + t y (0) = 1, y (0) = 1

y = c_{1} cos (t) + c_{2} sin (t) + 2

Solution steps

y^{''} (t) + y = 2

Solve linear ODE:

y = c_{1} cos (t) + c_{2} sin (t) + 2

y = c_{1} cos (t) + c_{2} sin (t) + 2

y^{^{''}} + 25 y = 8 sec (5 x)

y^{^{''}} - y = (1 - e^{2 x})^{- \frac{1}{2}}

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 8 y = 8 e^{- 2 x}

y^{^{''}} + y = 0.75 sin (t) - 0.25 sin (3 t)

y^{^{''}} + u^{2} y = cos (u t)

What is the solution for y^{''}+y=2,+ty(0)=1,y(0)=1 ?
The solution for y^{''}+y=2,+ty(0)=1,y(0)=1 is y=c_{1}cos(t)+c_{2}sin(t)+2