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,
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.
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?
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
$y(ln(x)+ 1)^2+ cy+ 2= 0$