Question

Find the general solution of the ode

xy'+y = (y^2)ln x

Answer 1

Are you sure it shouldn't be \(xy'-y=y^2\log x\) or \(-xy'+y=y^2\log x\)?

Answer 2

$$-xy'+y=y^2\log x\\-x\frac{y'}{y^2}+\frac1y=\log x$$Recognize our left-hand side is merely the product rule i.e. \((x/y)'=-xy'/y^2+y\):$$(x/y)'=\int\log x\,dx\\x/y=\int\log x\,dx=x\log x-x+C\\1/y=\log x-1+C/x\\y=\frac1{\log x-1+C/x}=\frac{x}{x\log x-x+C}$$

Answer 3

what i wrote is the problem... and I put it into WA. and it told me the solution was

y = 1 / (Cx + ln x + 1)

Answer 4

and this problem is in our separable section, but I do not see how to separate the x and y's

Answer 5

if it really is as you've stated then the problem is a tad bit different since it's actually a Bernoulli equation:$$xy'+y=y^2\log x\\-x\frac{y'}{y^2}-\frac1y=-\log x$$Let \(v=1/y\) so \(v'=-y'/y^2\):$$xv'-v=-\log x\\v'-\frac1xv=-\frac1x\log x$$Observe we can use an integration factor \(\mu=1/x\):$$\frac1xv'-\frac1{x^2}v=-\frac1{x^2}\log x$$again we note the product rule: \((v/x)'=v'/x-v/x^2\):$$(v/x)'=-\frac1{x^2}\log x\\\frac{v}x=-\int\frac1{x^2}\log x\,dx$$Using a substitution \(u=\log x\) we find \(du=\frac1x dx\) and we're left with:$$v=x\int -ue^{-u}\,du=x(ue^{-u}+e^{-u}+C)=x\left(\frac1x\log x+\frac1x+C\right)=1+\log x+Cx\\1/y=1+\log x+Cx\\y=\frac1{1+\log x+Cx}$$

Answer 6

to do this one by separation though... hmm. have you studied substitutions?

Answer 7

We did do bernoulli the section before but I just didnt see it. but once you said that i do see it now. i could just divide the whole thing by x, and n = 2. thank you for the help.

Answer 8

no problem! the reason it took me so long is because I kept messing up that last integral