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6.50 Another example

Example.

Find the solution of the initial-value problem

d2⁢yd⁢x2+α2⁢y=cos⁡α⁢x, y⁢(0)=y′⁢(0)=0\frac{d^{2}y}{dx^{2}}+\alpha^{2}y=\cos\alpha x,\;\;y(0)=y^{\prime}(0)=0

where α\alpha is a non-zero constant.

We apply the Laplace transform to both sides. We have

ℒ⁢(y′′)=-y′⁢(0)+s⁢ℒ⁢(y′)=s⁢ℒ⁢(y′){\mathcal{L}}(y^{\prime\prime})=-y^{\prime}(0)+s{\mathcal{L}}(y^{\prime})=s{% \mathcal{L}}(y^{\prime})

and similarly, ℒ⁢(y′)=s⁢ℒ⁢(y){\mathcal{L}}(y^{\prime})=s{\mathcal{L}}(y). Thus ℒ⁢(y′′+α2⁢y)=(s2+α2)⁢ℒ⁢(y){\mathcal{L}}\left(y^{\prime\prime}+\alpha^{2}y\right)=(s^{2}+\alpha^{2}){% \mathcal{L}}(y).

On the other side, we have ℒ⁢(cos⁡α⁢x)=ss2+α2.{\mathcal{L}}(\cos\alpha x)={\frac{s}{s^{2}+\alpha^{2}}.} Equating both sides and dividing by (s2+α2)(s^{2}+\alpha^{2}), we have ℒ⁢(y)=s(s2+α2)2.{\mathcal{L}}(y)={\frac{s}{(s^{2}+\alpha^{2})^{2}}.}