find the gen. solution of dy/dx + ylogy = ye^x
using the substituion u = logy
thankyou!
Hi
Let's do the substitution : $\displaystyle u=\ln y \Leftrightarrow y=\exp u\implies \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}u} {\mathrm{d}x}\exp u$
Substitute $\displaystyle y=\exp u$ and $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}u} {\mathrm{d}x}\exp u$ in the equation : $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}+y\ln y=y\exp x \Leftrightarrow \frac{\mathrm{d}u}{\mathrm{d}x}\exp u+u \exp u = \exp u \exp x$
Simplify by $\displaystyle \exp u$ and the equation becomes $\displaystyle \frac{\mathrm{d}u}{\mathrm{d}x}+u =\exp x$
This is a first order ODE hence you've to solve the homogeneous equation first : $\displaystyle \frac{\mathrm{d}u}{\mathrm{d}x}+u =0$. It'll give a solution $\displaystyle u_H$. Then, you need to find a particular solution $\displaystyle u_P$ of $\displaystyle E$. As the second member is $\displaystyle \exp x$, you may try $\displaystyle u_P(x)=\lambda \exp x$. (find $\displaystyle \lambda$ by replacing $\displaystyle u_P$ by its expression in $\displaystyle E$) The solutions of $\displaystyle E$ will be given by the sum $\displaystyle u=u_P+u_H$ and then you can use $\displaystyle y=\ln u$ to get the solutions in terms of $\displaystyle y$.
The first thing to do is to solve the homogeneous equation : $\displaystyle \frac{\mathrm{d}u}{\mathrm{d}x}+u=0 \Leftrightarrow\frac{\mathrm{d}u}{\mathrm{d}x}=-u \implies$ the solutions are $\displaystyle u_H(x)=K\exp(- x)$
Then, we have to find a particular solution $\displaystyle u_P$ of $\displaystyle \frac{\mathrm{d}u}{\mathrm{d}x}+u=\exp x$. Using my own advice, I try with $\displaystyle u_P(x)=\lambda \exp x \implies \frac{\mathrm{d}u_P}{\mathrm{d}x}=\lambda\exp x$. Let's replace this in the equation and solve for $\displaystyle \lambda$ :
$\displaystyle \frac{\mathrm{d}u_P}{\mathrm{d}x}+u_P=\exp x \implies \lambda\exp x+\lambda\exp x=\exp x$
Simplifying both sides by $\displaystyle \exp x$, we get $\displaystyle \lambda=\frac{1}{2}$ whence $\displaystyle u_P(x)=\frac{1}{2}\exp x$
The solutions of the ODE in terms of $\displaystyle u$ are given by $\displaystyle \boxed{u(x)=u_H(x)+u_P(x)=K\exp(-x)+\frac{1}{2}\exp x}$
Using $\displaystyle y=\exp u$, we get the solutions of $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} + y\ln y = y\exp x$ : $\displaystyle \boxed{y(x)=\exp\left(K\exp(-x)+\frac{1}{2}\exp x\right)}$
Also asked by a different member here: http://www.mathhelpforum.com/math-he...tion-help.html
and replied too.