# Give the correct answer and solve the given equation xy′′ − y′ = (3 + x)x^{2}e^{x}

Give the correct answer and solve the given equation $xy\prime \prime -y\prime =\left(3+x\right){x}^{2}{e}^{x}$
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krolaniaN

Given: consider the differential equation
$xy"-{y}^{\prime }=\left(3+x\right){x}^{2}{e}^{x}$
$⇒\frac{xy-{y}^{\prime }}{{x}^{2}}=\left(3+x\right){e}^{x}$
$⇒d\left(\frac{{y}^{\prime }}{x}\right)=\left(2+x\right){e}^{x}dx$
$⇒\int d\left(\frac{{y}^{\prime }}{x}\right)=\int \left(3+x\right){e}^{x}dx+C$
C is the integrating constant
$⇒\frac{{y}^{\prime }}{x}=3{e}^{x}+x{e}^{x}-{e}^{x}+$
$⇒{y}^{\prime }=2x{e}^{x}+x62{e}^{x}+Cx$
$⇒\frac{dy}{dx}=2x{e}^{x}+{x}^{2}{e}^{x}+Cx$
$⇒\int dy=2\int x{e}^{x}dx+\int {x}^{2}{e}^{x}dx+C\int xdx+{C}_{2}$
${C}_{2}$ is the integrating constant
$⇒y=2\left(x{e}^{x}-{e}^{x}\right)+{x}^{2}{e}^{x}-2x{e}^{x}+xe6x+C\frac{{x}^{2}}{2}+{C}_{2}$
$⇒$ $y={x}^{2}{e}^{x}+{X}_{1}{x}^{2}+{c}_{2}$
Where ${c}_{1}=\frac{C}{2}$ is the integrating constant
Finally result:
$⇒y={x}^{2}{e}^{x}+{X}_{1}{x}^{2}+{c}_{2}$