If frac{d^{2} y}{d x^{2}}=sin x+e^{x},y(0)=3, | vedclass

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(C) Given $\frac{d^{2} y}{d x^{2}}=\sin x+e^{x}$.Integrating both sides with respect to $x$:$\frac{d y}{d x} = \int (\sin x + e^{x}) dx = -\cos x + e^

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$y=2+4 x+e^{x}-\sin x$

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(C) Given $\frac{d^{2} y}{d x^{2}}=\sin x+e^{x}$.Integrating both sides with respect to $x$:$\frac{d y}{d x} = \int (\sin x + e^{x}) dx = -\cos x + e^{x} + c_{1}$.Given $\frac{d y}{d x} = 4$ at $x=0$:$4 = -\cos(0) + e^{0} + c_{1} \implies 4 = -1 + 1 + c_{1} \implies c_{1} = 4$.So,$\frac{d y}{d x} = e^{x} - \cos x + 4$.Integrating both sides again with respect to $x$:$y = \int (e^{x} - \cos x + 4) dx = e^{x} - \sin x + 4x + c_{2}$.Given $y(0) = 3$:$3 = e^{0} - \sin(0) + 4(0) + c_{2} \implies 3 = 1 - 0 + 0 + c_{2} \implies c_{2} = 2$.Thus,the equation of the curve is $y = e^{x} - \sin x + 4x + 2$.

If $\frac{d^{2} y}{d x^{2}}=\sin x+e^{x}$,$y(0)=3$,and $\frac{d y}{d x}$ at $x=0$ is $4$,then find the equation of the curve.

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