I'm practicing for the DE midterm, going through (supposedly) basic questions to practice.
I am now stuck on this question. I am supposed to find the general equation for the given DE. I don't know what I did wrong. I think the approach is correct, but possibly a calculation error? I can't quite figure it out (I've been trying it for at least an hour)
I divided $x^2$ to get rid of the coefficient on $y'$, and then applied Bernoulli's method. Next, I used the intergating factor method since it seemed appropriate to use. Some calculation followed and I got y(x) = C*x^(9/2) - which not an answer.
Can anyone point out where I got it wrong? Sorry for the bad handwriting in advance.$$x^2y'+2xy=5y^4$$
3 Answers
$\begingroup$$$x^2y'+2xy=5y^4$$$$x^2\frac{y'}{y^4}+2x\frac{1}{y^3}=5$$$$x^2\frac{-1}{3}\left(\frac{1}{y^3}\right)'+2x\frac{1}{y^3}=5$$Obviously the change of function $v=\frac{1}{y^3}$ is better than $v=y^3$.$$v=\frac{1}{y^3}\quad\text{leads to}\quad -x^2v'+6xv=15$$which is a linear ODE, easy to solve :$$v=cx^6+\frac{15}{7x}$$$$y^3=\frac{1}{cx^6+\frac{15}{7x}}$$
$\endgroup$ 2 $\begingroup$Hint
Being almost blind, it is quite difficult to me to read your notes.
From what I can see, you used $y={v^3}$; use instead $y^3=\frac 1v$ and you will arrive to something simple since the equation will become$$x^2 v'-6 x v+15=0$$
$\endgroup$ 0 $\begingroup$$$x^2y'+2xy=5y^4$$It's really hard to read the picture ...and to point out where you made mistakes$$(x^2y)'=\frac {5y^4x^8} {x^8}$$$$\int \frac {dx^2y}{x^8y^4}=5\int \frac {dx}{x^8}$$$$\int \frac {dx^2y}{(x^2y)^4}=- \frac {5}{7x^7}+K$$$$ \frac {1}{3(x^2y)^3}= \frac {5}{7x^7}+K$$$$ x= y^3(\frac {15}{7}+Cx^7)$$$$\implies y^3= \frac x {(\frac {15}{7}+Cx^7)}$$
$\endgroup$
