# Math Help - Substitution question

1. ## Substitution question

I've been stuck on this problem all day and can't seem get my solution to match the one in my textbook.

Use the appropriate method of substitution to solve the equation:

$xdx + (y-2x)dy = 0$

The xdx part is giving me some trouble. I tried using u = y/x but I only got to this point:

$\int dx/x = \int 1/(u-1)du$

$(x-y)ln|x-y| = y + c(x-y)$

2. Originally Posted by mdmac
I've been stuck on this problem all day and can't seem get my solution to match the one in my textbook.

Use the appropriate method of substitution to solve the equation:

$xdx + (y-2x)dy = 0$

The xdx part is giving me some trouble. I tried using u = y/x but I only got to this point:

$\int dx/x = \int 1/(u-1)du$

$(x-y)ln|x-y| = {\color{red}y}\, + c(x-y)$
First, divide the equation through by x to get $\,dx +\left(\frac{y}{x}-2\right)\,dy=0$

Now applying the substitution $y=ux\implies \,dy=u\,dx+x\,du$, we see that the DE becomes

$\,dx+\left(u-2\right)\left(u\,dx+x\,du\right)=0\implies \,dx+u^2\,dx+ux\,du-2u\,dx-2x\,du=0$ $\implies -\left(u^2-2u+1\right)\,dx=\left(u-2\right)x\,du$

Now separating the variables, we have $-\frac{\,dx}{x}=\frac{u-2}{\left(u-1\right)^2}\,du$

Now we integrate. The LHS becomes $-\ln x + C$.

Now we need to do some manipulation on the RHS.

Note that $\frac{u-2}{\left(u-1\right)^2}=\frac{u-1}{\left(u-1\right)^2}-\frac{1}{\left(u-1\right)^2}=\frac{1}{u-1}-\frac{1}{\left(u-1\right)^2}$. Now when we integrate wrt u, we have $\ln\!\left|u-1\right|+\frac{1}{u-1}+C$.

So it turns out that we have $-\ln x + C=\ln\!\left|u-1\right|+\frac{1}{u-1}+C\implies -\ln x= \ln\!\left|u-1\right|+\frac{1}{u-1}+K$.

Backsubstituting $u=\frac{y}{x}$, we have $-\ln x=\ln\!\left|\frac{y}{x}-1\right|+\frac{1}{\frac{y}{x}-1}+K\implies -\ln x=\ln\!\left|y-x\right|-\ln x+\frac{x}{y-x}+K$

Continuing with the simplification, we have $\frac{x}{x-y}+K=\ln\!\left|x-y\right|\implies {\color{red}x}\,+K\left(x-y\right)=\left(x-y\right)\ln\!\left|x-y\right|$

Hmm...

My solution and the books solution disagree...due to one variable...XD

Are you sure you copied it down correctly?

3. Originally Posted by Chris L T521
First, divide the equation through by x to get $\,dx +\left(\frac{y}{x}-2\right)\,dy=0$

Now applying the substitution $y=ux\implies \,dy=u\,dx+x\,du$, we see that the DE becomes

$\,dx+\left(u-2\right)\left(u\,dx+x\,du\right)=0\implies \,dx+u^2\,dx+ux\,du-2u\,dx-2x\,du=0$ $\implies -\left(u^2-2u+1\right)\,dx=\left(u-2\right)x\,du$

Now separating the variables, we have $-\frac{\,dx}{x}=\frac{u-2}{\left(u-1\right)^2}\,du$

Now we integrate. The LHS becomes $-\ln x + C$.

Now we need to do some manipulation on the RHS.

Note that $\frac{u-2}{\left(u-1\right)^2}=\frac{u-1}{\left(u-1\right)^2}-\frac{1}{\left(u-1\right)^2}=\frac{1}{u-1}-\frac{1}{\left(u-1\right)^2}$. Now when we integrate wrt u, we have $\ln\!\left|u-1\right|+\frac{1}{u-1}+C$.

So it turns out that we have $-\ln x + C=\ln\!\left|u-1\right|+\frac{1}{u-1}+C\implies -\ln x= \ln\!\left|u-1\right|+\frac{1}{u-1}+K$.

Backsubstituting $u=\frac{y}{x}$, we have $-\ln x=\ln\!\left|\frac{y}{x}-1\right|+\frac{1}{\frac{y}{x}-1}+K\implies -\ln x=\ln\!\left|y-x\right|-\ln x+\frac{x}{y-x}+K$

Continuing with the simplification, we have $\frac{x}{x-y}+K=\ln\!\left|x-y\right|\implies {\color{red}x}\,+K\left(x-y\right)=\left(x-y\right)\ln\!\left|x-y\right|$

Hmm...

My solution and the books solution disagree...due to one variable...XD

Are you sure you copied it down correctly?
I just doubled checked and the book shows a y rather than an x. Your explanation so far has been extremely helpful though.

4. Originally Posted by mdmac
I just doubled checked and the book shows a y rather than an x. Your explanation so far has been extremely helpful though.
I checked my solution through another means, and it should be x, not y. So maybe its a typo in your book...

5. Originally Posted by Chris L T521
I checked my solution through another means, and it should be x, not y. So maybe its a typo in your book...
I really does seem that way. Man, what a lousy deal. Thank you for your help though, I was pulling my hair out!