Solve differential equation y'+3x^2y= \sin(x)e^{-x^3}, \ y(0)=1

Solve differential equation
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${e}^{\int pdx}={e}^{\int 3{x}^{2}dx}$
${e}^{\int pdx}={e}^{\frac{3{x}^{3}}{3}}$
${e}^{\int pdx}={e}^{{x}^{3}}$
That is the I.F. is $\mu \left(x\right)={e}^{{x}^{3}}$
Multiply the integrating factor on both sides of the given equation
${e}^{{x}^{3}}{y}^{\prime }=\mathrm{sin}\left(x\right){e}^{-{x}^{3}}{e}^{{x}^{3}}$
${e}^{{x}^{3}}{y}^{\prime }=\mathrm{sin}\left(x\right)\left[\because {e}^{-{x}^{3}}{e}^{{x}^{3}}={e}^{-{x}^{3}+{x}^{3}}=1\right]$
Integrate both sides of the above equation
$\int \frac{d}{dx}\left({e}^{{x}^{3}}y\right)dx=\int \mathrm{sin}\left(x\right)dx$
${e}^{{x}^{3}}y=-\mathrm{cos}\left(x\right)+c$
$y=\frac{-\mathrm{cos}\left(x\right)+c}{{e}^{{x}^{3}}}$
Now compute the value of constant C by applying the given condition as follows
$y\left(0\right)=\frac{-\mathrm{cos}\left(0\right)+c}{{e}^{{\left(0\right)}^{3}}}$

$1=-1+c$
$1+1=c$
$2=c$
Now substitute the value of C for the value of y and simplify further
$y=\frac{-\mathrm{cos}\left(x\right)+2}{{e}^{{x}^{3}}}$