# Thread: Stuck on two differential equations

1. ## Stuck on two differential equations

Hi,

The first equation is $dy + ln x \, y \, dx = (4x +lny) \, dx$. It doesn't seem to matter how I move around the variables, I can't group the y terms with dy and the x terms with dx.

The second equation is $x \frac{dy}{dx} = y^2 + y^2 \, lnx$. For this one I rearranged the terms to get $\frac{1}{y^2}dy = \frac{1}{x}dx + \frac{lnx}{x} \, dx$. Solving the integrals gave me $-\frac{1}{y} = lnx + \frac{1}{2}(lnx)^2$. I don't see the problem with this but the answer key gave me something different. Have I made an error?

Thanks.

2. ## Re: Stuck on two differential equations

for the 2nd equation you end up with

$-\dfrac 1 y = \ln(x)+\dfrac 1 2 (\ln(x))^2 + C_1$

$y = -\dfrac 1 {2\ln(x) + (\ln(x))^2 + C}$, $C=2 C_1$

still looking at the first one

could you post the original form of the first diff eq?

3. ## Re: Stuck on two differential equations

One obvious point is that you have forgotten the "constant of integration". But was the "something different" in the answer key?

4. ## Re: Stuck on two differential equations

Originally Posted by romsek
for the 2nd equation you end up with

$-\dfrac 1 y = \ln(x)+\dfrac 1 2 (\ln(x))^2 + C_1$

$y = -\dfrac 1 {2\ln(x) + (\ln(x))^2 + C}$, $C=2 C_1$

still looking at the first one

could you post the original form of the first diff eq?
Hi Romsek, the equation I posted is the original form straight from the textbook.

Originally Posted by HallsofIvy
One obvious point is that you have forgotten the "constant of integration". But was the "something different" in the answer key?
Hi HallsOfIvy, the answer key gives $y(1 + \ln x)^2 + cy + 2 = 0$.

5. ## Re: Stuck on two differential equations

Originally Posted by KevinShaughnessy
Hi Romsek, the equation I posted is the original form straight from the textbook.

Hi HallsOfIvy, the answer key gives $y(1 + \ln x)^2 + cy + 2 = 0$.
Okay, you had $-\frac{1}{y}= ln(x)+ \frac{1}{2}(ln(x))^2+ C$ where "C" is the constant of integration. An obvious thing to do is to multiply through by 2y: $-2= 2yln(x)+ y(ln(x))^2+ 2Cy$

That is the same as $y((ln(x))^2+ 2ln(x))+ 2Cy+ 2= 0$

Complete the square! $y((ln(x))^2+ 2ln(x)+ 1- 1)+ 2Cy+ 2= y(ln(x)+ 1)^2- y+ 2Cy+ 2= 0$

$y(ln(x)+ 1)^2+ (2C- 1)y+ 2= 0$

Finally, since "C" is an undetermined constant, so is "2C- 1". Let c= 2C- 1 so that we have
$y(ln(x)+ 1)^2+ cy+ 2= 0$