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

The solution for y^{''}+y=cos(x),y(0)=1,y^'(0)=-1 is y=cos(x)-sin(x)+(xsin(x))/2

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

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

y = cos (x) - sin (x) + \frac{x sin ( x )}{2}

Solution steps

y^{''} (x) + y = cos (x)

Solve linear ODE:

y = cos (x) - sin (x) + \frac{x sin ( x )}{2}

y = cos (x) - sin (x) + \frac{x sin ( x )}{2}

y^{^{''}} + 5 y^{^{'}} + 4 y = 8

y^{^{''}} - 100 y = cosh (10 x)

y^{^{''}} - y^{^{'}} + 5 y = e^{2 x}

y^{^{''}} - 2 y^{^{'}} + y = 8 e^{x} - 5

\frac{d ^{2} y}{d x ^{2}} + 8 \frac{d y}{d x} + 16 y = 16 sin (x)

What is the solution for y^{''}+y=cos(x),y(0)=1,y^'(0)=-1 ?
The solution for y^{''}+y=cos(x),y(0)=1,y^'(0)=-1 is y=cos(x)-sin(x)+(xsin(x))/2