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Question Number 217300 by peter frank last updated on 09/Mar/25

Answered by Marzuk last updated on 09/Mar/25

$$\frac{d^{2}y}{{dx}^2}-2a\frac{dy}{dx}+(a^2+b^2)y=0$$$$or,\frac{{dy}^{\prime }}{dx}-2a\frac{d}{dx}\left[e^{ax}sin\left(bx\right)\right]+a^{2}y+b^{2}y=0$$$$Here:\frac{{dy}^{\prime }}{dx}=\frac{d}{dx}\left(\frac{d}{dx}\left[e^{ax}sin\left(bx\right)\right]\right)$$$$\frac{d}{dx}\left[e^{ax}sin\left(bx\right)\right]={ae}^{ax}sin\left(bx\right)+e^{ax}bcos\left(bx\right)$$$$\therefore \frac{d}{dx}[{ae}^{ax}sin\left(bx\right)+e^{ax}bcos\left(bx\right)]$$$$=\frac{d}{dx}\left[{ae}^{ax}sin\left(bx\right)\right]+\frac{d}{dx}\left[e^{ax}bcos\left(bx\right)\right]$$$$=a^2{e}^{ax}sin\left(bx\right)+{ae}^{ax}bcos\left(bx\right)+{ae}^{ax}bcos\left(bx\right)-b^{2}sin\left(bx\right)$$$$or,a^2{e}^{ax}sin\left(bx\right)+{ae}^{ax}bcos\left(bx\right)+{ae}^{ax}bcos\left(bx\right)-b^{2}sin\left(bx\right)-2a^2{e}^{ax}sin\left(bx\right)-2{ae}^{ax}bcos\left(bx\right)+a^2{e}^{ax}sin\left(bx\right)+b^2{e}^{ax}sin\left(bx\right)=0$$$$or,a^2{e}^{ax}sin\left(bx\right)+a^2{e}^{ax}sin\left(bx\right)-2a^2{e}^{ax}sin\left(bx\right)+{ae}^{ax}bcos\left(bx\right)+{ae}^{ax}bcos\left(bx\right)-2{ae}^{ax}bcos\left(bx\right)+b^2{e}^{ax}sin\left(bx\right)+b^2{e}^{ax}sin\left(bx\right)-2b^2{e}^{ax}sin\left(bx\right)=0$$$$or,0+0+0=0$$$$or,0=0$$$$\therefore \frac{d^{2}y}{{dx}^2}-2a\frac{dy}{dx}+(a^2+b^2)y=0$$

Answered by som(math1967) last updated on 10/Mar/25

$$y=e^{\mathit{a}\mathit{x}}\mathit{s}\mathit{i}\mathit{n}\mathit{b}\mathit{x}$$$${\mathit{y}}_1={\mathit{a}\mathit{e}}^{\mathit{a}\mathit{x}}\mathit{s}\mathit{i}\mathit{n}\mathit{b}\mathit{x}+{\mathit{b}\mathit{e}}^{\mathit{a}\mathit{x}}\mathit{c}\mathit{o}\mathit{s}\mathit{b}\mathit{x}$$$${\mathit{y}}_1=\mathit{a}\mathit{y}+{\mathit{b}\mathit{e}}^{\mathit{a}\mathit{x}}\mathit{c}\mathit{o}\mathit{s}\mathit{b}\mathit{x}[\mathit{y}={\mathit{e}}^{\mathit{a}\mathit{x}}\mathit{s}\mathit{i}\mathit{n}\mathit{b}\mathit{x}[$$$${\mathit{y}}_2={\mathit{a}\mathit{y}}_1+{\mathit{a}\mathit{b}\mathit{e}}^{\mathit{a}\mathit{x}}\mathit{c}\mathit{o}\mathit{s}\mathit{b}\mathit{x}-{\mathit{b}}^2{\mathit{e}}^{\mathit{a}\mathit{x}}\mathit{s}\mathit{i}\mathit{n}\mathit{b}\mathit{x}$$$${\mathit{y}}_2={\mathit{a}\mathit{y}}_1+\mathit{a}({\mathit{y}}_1-\mathit{a}\mathit{y})-{\mathit{b}}^2\mathit{y}$$$$[{\mathit{b}\mathit{e}}^{\mathit{a}\mathit{x}}\mathit{c}\mathit{o}\mathit{s}\mathit{b}\mathit{x}={\mathit{y}}_1-\mathit{a}\mathit{y}]$$$${\mathit{y}}_2-2{\mathit{a}\mathit{y}}_1+({\mathit{a}}^2+{\mathit{b}}^2)\mathit{y}=0$$$$[{\mathit{y}}_1=\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}{\mathit{y}}_2=\frac{{\mathit{d}}^2\mathit{y}}{{\mathit{d}\mathit{x}}^2}]$$

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