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Question Number 133776 by bramlexs22 last updated on 24/Feb/21

$$\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=\sin 3\mathrm{x}+{\mathrm{e}}^{\mathrm{x}}+{\mathrm{x}}^2,\mathrm{when}{\mathrm{y}}^{\prime }\left(0\right)=1$$$$\mathrm{and}\mathrm{y}\left(0\right)=0.\mathrm{Find}\mathrm{the}\mathrm{solution}$$

Answered by bobhans last updated on 24/Feb/21

$$\frac{d}{dx}\left[\frac{dy}{dx}\right]=\sin 3x+e^x+x^2$$$$d\left[\frac{dy}{dx}\right]=(\sin 3x+e^x+x^2)dx$$$$\int d\left[\frac{dy}{dx}\right]=\int (\sin 3x+e^x+x^2)dx$$$$\Rightarrow \frac{dy}{dx}=-\frac{1}{3}\cos 3x+e^x+\frac{1}{3}{x}^3+C_1$$$$y^{\prime }\left(0\right)=-\frac{1}{3}+1+C_1=1;C_1=\frac{1}{3}$$$$$$$$\int dy=\int (-\frac{1}{3}\cos 3x+e^x+\frac{1}{3}{x}^3+\frac{1}{3})dx$$$$y=-\frac{1}{9}\sin 3x+e^x+\frac{1}{12}{x}^4+\frac{1}{3}x+C_2$$$$y\left(0\right)=1+C_2=0\Rightarrow C_2=-1$$$$$$$$\therefore y=-\frac{\sin 3x}{9}+e^x+\frac{x^4+4x-12}{12}$$

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