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alina1380 [7]
3 years ago
8

Find the solution of the differential equation that satisfies the given initial condition. dy/dx = xe^y, y(0) = 0

Mathematics
1 answer:
Anna [14]3 years ago
3 0
\dfrac{\mathrm dy}{\mathrm dx}=xe^y
e^{-y}\,\mathrm dy=x\,\mathrm dx
-e^{-y}=\dfrac{x^2}2+C

y(0)=0
\implies-1=C
\implies -e^{-y}=\dfrac{x^2}2-1
e^{-y}=1-\dfrac{x^2}2
-y=\ln\left(1-\dfrac{x^2}2\right)
y=-\ln\left(1-\dfrac{x^2}2\right)
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