# calculus - homogenous de

• May 11th 2008, 02:11 AM
mathshelpbook
calculus - homogenous de
find the gen. solution of dy/dx + ylogy = ye^x
using the substituion u = logy

thankyou!
• May 11th 2008, 03:25 AM
flyingsquirrel
Hi

Let's do the substitution : $u=\ln y \Leftrightarrow y=\exp u\implies \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}u} {\mathrm{d}x}\exp u$

Substitute $y=\exp u$ and $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}u} {\mathrm{d}x}\exp u$ in the equation : $\frac{\mathrm{d}y}{\mathrm{d}x}+y\ln y=y\exp x \Leftrightarrow \frac{\mathrm{d}u}{\mathrm{d}x}\exp u+u \exp u = \exp u \exp x$

Simplify by $\exp u$ and the equation becomes $\frac{\mathrm{d}u}{\mathrm{d}x}+u =\exp x$

This is a first order ODE hence you've to solve the homogeneous equation first : $\frac{\mathrm{d}u}{\mathrm{d}x}+u =0$. It'll give a solution $u_H$. Then, you need to find a particular solution $u_P$ of $E$. As the second member is $\exp x$, you may try $u_P(x)=\lambda \exp x$. (find $\lambda$ by replacing $u_P$ by its expression in $E$) The solutions of $E$ will be given by the sum $u=u_P+u_H$ and then you can use $y=\ln u$ to get the solutions in terms of $y$.
• May 11th 2008, 03:51 AM
mathshelpbook
thanks for your help! ok so i understand up to du/dx + u = e^x, but then i'm confused as to how to continue solving?
• May 11th 2008, 04:51 AM
flyingsquirrel
The first thing to do is to solve the homogeneous equation : $\frac{\mathrm{d}u}{\mathrm{d}x}+u=0 \Leftrightarrow\frac{\mathrm{d}u}{\mathrm{d}x}=-u \implies$ the solutions are $u_H(x)=K\exp(- x)$

Then, we have to find a particular solution $u_P$ of $\frac{\mathrm{d}u}{\mathrm{d}x}+u=\exp x$. Using my own advice, I try with $u_P(x)=\lambda \exp x \implies \frac{\mathrm{d}u_P}{\mathrm{d}x}=\lambda\exp x$. Let's replace this in the equation and solve for $\lambda$ :

$\frac{\mathrm{d}u_P}{\mathrm{d}x}+u_P=\exp x \implies \lambda\exp x+\lambda\exp x=\exp x$

Simplifying both sides by $\exp x$, we get $\lambda=\frac{1}{2}$ whence $u_P(x)=\frac{1}{2}\exp x$

The solutions of the ODE in terms of $u$ are given by $\boxed{u(x)=u_H(x)+u_P(x)=K\exp(-x)+\frac{1}{2}\exp x}$

Using $y=\exp u$, we get the solutions of $\frac{\mathrm{d}y}{\mathrm{d}x} + y\ln y = y\exp x$ : $\boxed{y(x)=\exp\left(K\exp(-x)+\frac{1}{2}\exp x\right)}$
• May 11th 2008, 08:04 AM
mr fantastic
Quote:

Originally Posted by mathshelpbook
find the gen. solution of dy/dx + ylogy = ye^x
using the substituion u = logy

thankyou!

Also asked by a different member here: http://www.mathhelpforum.com/math-he...tion-help.html

and replied too.