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

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

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

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

y = cos (t) - 2 sin (t) + t

Solution steps

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

Solve linear ODE:

y = cos (t) - 2 sin (t) + t

y = cos (t) - 2 sin (t) + t

\frac{\partial}{\partial y} (\frac{x}{x ^{7} - y ^{3}})

\frac{d}{d x} (\frac{3 x ^{2} + 1}{x + 2})

\int \frac{x + 2}{( 3 + 2 x - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{d}{d x} (a e^{b x})

x \to 2 lim (2 x)

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