# The variables x and y satisfy the differential equation dy/dx= xe^x+y. It is given that y = 0 when x

###### Question:

(ii) Explain why x can only take values that are less than 1.

## Answers

Rewrite the ODE

[tex]\dfrac{\mathrm dy}{\mathrm dx}=xe^x+y[/tex]

as

[tex]\dfrac{\mathrm dy}{\mathrm dx}-y=xe^x[/tex]

Divide both sides by [tex]e^x[/tex]:

[tex]e^{-x}\dfrac{\mathrm dy}{\mathrm dx}-e^{-x}y=x[/tex]

The left side condenses to the derivative of a product:

[tex]\dfrac{\mathrm d}{\mathrm dx}\left[e^{-x}y\right]=x[/tex]

Integrate both sides and solve for [tex]y[/tex]:

[tex]e^{-x}y=\displaystyle\int x\,\mathrm dx=\frac12x^2+C[/tex]

[tex]y=\dfrac12x^2e^x+Ce^x[/tex]

Given that [tex]y=0[/tex] when [tex]x=0[/tex], we find

[tex]0=\dfrac120^2e^0+Ce^0\implies C=0[/tex]

so that the solution is

[tex]\boxed{y(x)=\dfrac{x^2e^x}2}[/tex]

I'm not sure I understand the claim that [tex]x[/tex] must be less than 1...

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step-by-step explanation:

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1696 hundred thousand

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