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Find an explicit solution of the given initial-value problem.dy/dx = ye^x^2, y(3) = 1.

User K D
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1 Answer

5 votes

Answer:


y=exp(\int\limits^x_4 {e^{-t^(2) } } \, dt)

Explanation:

This is a separable equation with an initial value i.e. y(3)=1.

Take y from right hand side and divide to left hand side ;Take dx from left hand side and multiply to right hand side:


(dy)/(y) =e^{-x^(2) }dx

Take t as a dummy variable, integrate both sides with respect to "t" and substituting x=t (e.g. dx=dt):


\int\limits^x_3 {(1)/(y) } \, (dy)/(dt) dt=\int\limits^x_3 {e^{-t^(2) } } dt

Integrate on both sides:


ln(y(t))\left \{ {{t=x} \atop {t=3}} \right. =\int\limits^x_3 {e^{-t^(2) } } \, dt

Use initial condition i.e. y(3) = 1:


ln(y(x))-(ln1)=\int\limits^x_3 {e^{-t^(2) } } \, dt\\ln(y(x))=\int\limits^x_3 {e^{-t^(2) } } \, dt\\

Taking exponents on both sides to remove "ln":


y=exp (\int\limits^x_3 {e^{-t^(2) } } \, dt)

User Kirk Broadhurst
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