Mathematics, 13.06.2020 08:57 angelrenee2000
The variables x and y satisfy the differential equation dy/dx= xe^x+y. It is given that y = 0 when x = 0.
(i) Solve the differential equation, obtaining y in terms of x.
(ii) Explain why x can only take values that are less than 1.
Answers: 2
:
![\dfrac{\mathrm d}{\mathrm dx}\left[e^{-x}y\right]=x](/tpl/images/0685/0251/8c1c2.png)
:
when 
, we find
must be less than 1...
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The variables x and y satisfy the differential equation dy/dx= xe^x+y. It is given that y = 0 when x...
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