Hi, a question how can I get the exact solution of the following non-linear problem ??
\(\displaystyle u''+uu'-u=e^{2x} , \ \ \ \ u(0)=1 , \ \ \ u(1)=e. \)
Try the following to solve the homogeneous equation:
Putting \(\displaystyle v=u' \rightarrow{ u''= v \frac{dv}{du}}\) by separating variables \(\displaystyle -v-\ln|1-v|=\frac{u^2}{2}+c_{1}\)
so \(\displaystyle v=1+c_{2} e^{-v-\frac{u^2}{2}+c_{1}} \rightarrow{ u'=1+c_{2} e^{-u'-\frac{u^2}{2}+c_{1}} }\) and from here I don't know how to find \(\displaystyle u(x)\)
\(\displaystyle u''+uu'-u=e^{2x} , \ \ \ \ u(0)=1 , \ \ \ u(1)=e. \)
Try the following to solve the homogeneous equation:
Putting \(\displaystyle v=u' \rightarrow{ u''= v \frac{dv}{du}}\) by separating variables \(\displaystyle -v-\ln|1-v|=\frac{u^2}{2}+c_{1}\)
so \(\displaystyle v=1+c_{2} e^{-v-\frac{u^2}{2}+c_{1}} \rightarrow{ u'=1+c_{2} e^{-u'-\frac{u^2}{2}+c_{1}} }\) and from here I don't know how to find \(\displaystyle u(x)\)