# Thread: How to solve this differential equation?

1. ## How to solve this differential equation?

Hello,
How to solve this differential equation $\frac{dy}{dx}=\frac{x+4y-3}{2x+3y-5}$ If we assume $u(x)=\frac{y(x)-\frac15}{x-\frac{11}{5}}$ how to proceed further?

2. ## Re: How to solve this differential equation?

if we let $\displaystyle v=x-\frac{11}{5}$ and $\displaystyle w=y-\frac{1}{5}$

we get

$\displaystyle \frac{\text{dw}}{\text{dx}}=\frac{\text{dy}}{\text {dx}}=\frac{x+4y-3}{2x+3y-5}=\frac{v+4 w}{2 v+3 w}=\frac{1+4u}{2+3u}$

on the other hand $\displaystyle w=u v$ so

$\displaystyle \frac{\text{dw}}{\text{dx}}=u +v \frac{\text{du}}{\text{dx}}=u+\left(x-\frac{11}{5}\right)\frac{\text{du}}{\text{dx}}$

3. ## Re: How to solve this differential equation?

Originally Posted by Idea
if we let $\displaystyle v=x-\frac{11}{5}$ and $\displaystyle w=y-\frac{1}{5}$

we get

$\displaystyle \frac{\text{dw}}{\text{dx}}=\frac{\text{dy}}{\text {dx}}=\frac{x+4y-3}{2x+3y-5}=\frac{v+4 w}{2 v+3 w}=\frac{1+4u}{2+3u}$

on the other hand $\displaystyle w=u v$ so

$\displaystyle \frac{\text{dw}}{\text{dx}}=u +v \frac{\text{du}}{\text{dx}}=u+\left(x-\frac{11}{5}\right)\frac{\text{du}}{\text{dx}}$
Hello,
What would be the interval of validity for x and what would be the initial condition if we assume C=1.

4. ## Re: How to solve this differential equation?

We know that the initial value problem

$\displaystyle y'(x)=\frac{x+4y-3}{2x+3y-5}$ , $\displaystyle y\left(x_0\right)=y_0$

where $\displaystyle \left(x_0,y_0\right)$ is not on the line $\displaystyle 2x+3y-5=0$,

has a unique solution in some interval containing $\displaystyle x_0$