What is the solution for a^2*y+y^{''}=cos(b*t),y(0)=0,y^'(0)=0 ?

The solution for a^2*y+y^{''}=cos(b*t),y(0)=0,y^'(0)=0 is y=(cos(bt))/(a^2-b^2)-(cos(at))/(a^2-b^2)

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a^2y+y^{''}=cos(bt),y(0)=0,y^'(0)=0

a^{2} \cdot y + y^{^{''}} = cos (b \cdot t), y (0) = 0, y^{^{'}} (0) = 0

y = \frac{cos ( b t )}{a ^{2} - b ^{2}} - \frac{cos ( a t )}{a ^{2} - b ^{2}}

Solution steps

a^{2} \cdot y + y^{''} (t) = cos (b \cdot t)

Solve linear ODE:

y = \frac{cos ( b t )}{a ^{2} - b ^{2}} - \frac{cos ( a t )}{a ^{2} - b ^{2}}

y = \frac{cos ( b t )}{a ^{2} - b ^{2}} - \frac{cos ( a t )}{a ^{2} - b ^{2}}

y^{^{''}} - y^{^{'}} = 1 - x^{2}

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 e^{x}

y^{^{''}} + 8 y^{^{'}} + 16 y = \frac{3 e ^{- 4 x}}{x}

(d^{3} - 3 d^{2} - 10 d + 24) y = 2 x^{2} + 5

y^{^{''}} - y^{^{'}} + 2 = 0

What is the solution for a^2*y+y^{''}=cos(b*t),y(0)=0,y^'(0)=0 ?
The solution for a^2*y+y^{''}=cos(b*t),y(0)=0,y^'(0)=0 is y=(cos(bt))/(a^2-b^2)-(cos(at))/(a^2-b^2)