find the gen. solution of dy/dx + ylogy = ye^x

using the substituion u = logy

thankyou!

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- May 11th 2008, 01:11 AMmathshelpbookcalculus - homogenous de
find the gen. solution of dy/dx + ylogy = ye^x

using the substituion u = logy

thankyou! - May 11th 2008, 02:25 AMflyingsquirrel
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 $\displaystyle y$. - May 11th 2008, 02:51 AMmathshelpbook
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, 03:51 AMflyingsquirrel
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 $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} + y\ln y = y\exp x$ : $\displaystyle \boxed{y(x)=\exp\left(K\exp(-x)+\frac{1}{2}\exp x\right)}$ - May 11th 2008, 07:04 AMmr fantastic
Also asked by a different member here: http://www.mathhelpforum.com/math-he...tion-help.html

and replied too.