Two differential equations

How would I solve these differential equations?

$$y'+2y^2=\frac{6}{x^2}$$

I tried finding integral product but couldn't find its integral. And also tried to trasform into homogen equation.

and the second one is:

$$xe^{2y}y'+e^{2y}=\frac{\ln x}{x}$$

How can I start? Thanks.

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Well, I can figure out the second one. My guess was we could get the left side of the equation to look like

$$\frac d{dx}(f(x)e^{2y})=2f(x)e^{2y}y'+f'(x)e^{2y}$$

through the use of an integrating factor. So we have

$$\frac{f'(x)}{2f(x)}=\frac1x$$

$$\ln f(x)=2\ln x$$ $$f(x)=x^2$$

To get the equation into the proper form, multiply both sides by $2x$.

$$2x^2e^{2y}y'+2xe^{2y}=\frac d{dx}(x^2e^{2y})=2\ln x$$

I assume you can take this one from here?

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Great use of the integrating factor. (+1) –  Pragabhava Nov 18 '12 at 20:43
$x^2e^{2y}=2(x\ln x -x+ c)$ I see. –  lyme Nov 19 '12 at 2:04
can we get that "x" the integrating factor from $e^{\int\frac{M_y - N_x}{N}dx}=$$e^{\ln x}=x ? thank you. – lyme Nov 19 '12 at 2:11 For the first equation, take y = \frac{A}{x}, then$$ y' + 2 y^2 = -\frac{A}{x^2} + 2 \frac{A^2}{x^2} = \frac{6}{x^2} $$Hence$$ 2 A^2 - A - 6 = 0 $$and$$ A = \begin{cases} \hskip.5cm 2 \\ -\tfrac{3}{2} \end{cases} $$There are two solutions$$ y_1(x) = -\frac{3}{2 x} \mbox{ and } y_2(x) = \frac{2}{x} $$If you want to do a "formal" derivation of the ansat, you should note that the first equation is a Riccati equation. - Thank you. if the equation was like y'+ay^2=bx^c, (a,b,c \in \Bbb R) would it be riccati again? – lyme Nov 19 '12 at 2:24 @mehmtDemir Indeed it is. A Ricatti equation is of the form$$y' = q_0(x) + q_1(x)y + \color{red}{q_2(x)y^2}.$$If it involves$y'$,$y$and$y^2\$ linearly it's a Riccati type equation, and can always be linearized. –  Pragabhava Nov 19 '12 at 2:35