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

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

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

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

y = - cos (t) + sin (t) + 1

Solution steps

y^{''} (t) + y = 1

Solve linear ODE:

y = - cos (t) + sin (t) + 1

y = - cos (t) + sin (t) + 1

y^{^{''}} + 2 y^{^{'}} + y = 4000 sec (40 x)

a x^{^{''}} + b x^{^{'}} + c = 0

\frac{d ^{2} y}{d t ^{2}} - 3 \frac{d y}{d t} - 10 y = 4 e^{5 t}

y^{^{''}} - y^{^{'}} - 12 y = e^{2} x + e^{x} - 2

y^{^{''}} + 6 y^{^{'}} + 5 y = 36 e^{x}, y (0) = - 2, y^{^{'}} (0) = 4

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