If y-x2ex- show that d2ydx2-dydx-2x-1ex-0

y = x^2e^x ⇒dydx = 2xe^x + x^2e^x = 2xe^x + y ⇒d^2ydx^2 = 2( e^x + xe^x) + dydx ⇒d^2ydx^2 dydx 2( x + 1)e^x = 0 ...

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Byju's Answer

Standard XII

Mathematics

Higher Order Derivatives

If y=x2ex, ...

Question

If $y = x^{2} e^{x}$ , show that $\frac{d^{2} y}{d x^{2}} - \frac{d y}{d x} - 2 (x + 1) e^{x} = 0$

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Solution

$y = x^{2} e^{x}$

$\Rightarrow \frac{d y}{d x} = 2 x e^{x} + x^{2} e^{x} = 2 x e^{x} + y$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = 2 (e^{x} + x e^{x}) + \frac{d y}{d x}$

$\Rightarrow \frac{d^{2} y}{d x^{2}} - \frac{d y}{d x} - 2 (x + 1) e^{x} = 0$

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If y=x2ex, show thatd2ydx2−dydx−2(x+1)ex=0

y=x2ex⇒dydx=2xex+x2ex=2xex+y⇒d2ydx2=2(ex+xex)+dydx⇒d2ydx2−dydx−2(x+1)ex=0

If $y = x^{2} e^{x}$ , show that $\frac{d^{2} y}{d x^{2}} - \frac{d y}{d x} - 2 (x + 1) e^{x} = 0$

$y = x^{2} e^{x}$

$\Rightarrow \frac{d y}{d x} = 2 x e^{x} + x^{2} e^{x} = 2 x e^{x} + y$

$\Rightarrow \frac{d^{2} y}{d x^{2}} = 2 (e^{x} + x e^{x}) + \frac{d y}{d x}$

$\Rightarrow \frac{d^{2} y}{d x^{2}} - \frac{d y}{d x} - 2 (x + 1) e^{x} = 0$