Answer:
[tex]y=exp(\int\limits^x_4 {e^{-t^{2} } } \, dt)[/tex]
Step-by-step 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:
[tex]\frac{dy}{y} =e^{-x^{2} }dx[/tex]
Take t as a dummy variable, integrate both sides with respect to "t" and substituting x=t (e.g. dx=dt):
[tex]\int\limits^x_3 {\frac{1}{y} } \, \frac{dy}{dt} dt=\int\limits^x_3 {e^{-t^{2} } } dt[/tex]
Integrate on both sides:
[tex]ln(y(t))\left \{ {{t=x} \atop {t=3}} \right. =\int\limits^x_3 {e^{-t^{2} } } \, dt[/tex]
Use initial condition i.e. y(3) = 1:
[tex]ln(y(x))-(ln1)=\int\limits^x_3 {e^{-t^{2} } } \, dt\\ln(y(x))=\int\limits^x_3 {e^{-t^{2} } } \, dt\\[/tex]
Taking exponents on both sides to remove "ln":
[tex]y=exp (\int\limits^x_3 {e^{-t^{2} } } \, dt)[/tex]
