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

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

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

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

y = x sin (x) + cos (x)

Solution steps

y^{''} (x) + y = 2 cos (x)

Solve linear ODE:

y = x sin (x) + cos (x)

y = x sin (x) + cos (x)

y^{^{''}} - 3 y^{^{'}} + y = 2 sin (2 x)

(D + 1)^{3} y = 16 (2 x + 3)^{- 1} e^{x}

y^{^{''}} - 4 y^{^{'}} - 8 y = 8 x^{2} + 8 x + 1

y^{^{''}} - 9 y = 63

D, D (x) = e^{2 x^{3} + 4}

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