Help solving the non-linear ODE $\frac{dy}{dt} = t^3 (y-t)^2 + \frac{y}{t}$

The particular solution given in the question is $$u(t) = t$$. I know that this is a Ricatti differential equation, therefore it's solution is $$y=u(t)+\frac{1}{v(t)}$$. So first I've write the ODE in the Ricatti form:

$$\frac{dy}{dt} = t^3 (y-t)^2 + \frac{y}{t}$$$$\frac{dy}{dt} = y^2(t^3)+y(-2t^4+\frac1t) +t^2$$

So I've used the substituition $$y=t+\frac{1}{v(t)} \Rightarrow y' = 1-\frac{dv}{v^2dt}$$.

$$t^3y^2 + y(-2t^4+\frac1t )+t^2 = 1 - \frac{dv}{v^2dt}$$ $$...$$ In the end, I've got: $$-t^5+\frac{1}{v^2}+\frac{1}{vt}=-\frac{dv}{v^2dt}$$ $$-t^5v^2+1+\frac{v}{t}=-\frac{dv}{dt}$$

But, the final equation would be a first order linear ODE, and this one isn't.

Can you help me solving this?

-- PS: Should I try to solve this new equation again using Ricatti? I will try to do this now. But, if you have any suggestions, please tell me

• Why did you expand and not just insert $(y-t)=\frac1v$? Somehow you missed a $t^5$ term. – Lutz Lehmann Oct 10 '19 at 15:05
• Oh, thank you! I will try to solve without expanding. – João Pedro Oct 10 '19 at 15:15

After change $$y=v(t)+t$$ we get Bernoulli equation $$\frac{d}{d t} \operatorname{v}(t)-\frac{\operatorname{v}(t)}{t}={{t}^{3}}\, {{\operatorname{v}(t)}^{2}}$$ You can continue.
$$\frac{dy}{dt} = t^3 (y-t)^2 + \frac{y}{t}$$ $$y(t)=u(t)+t$$ $$\frac{du}{dt} = t^3 u^2 + \frac{u}{t}$$ $$-\frac{1}{u^2}\frac{du}{dt} =- t^3 - \frac{1}{tu}$$ $$v(t)=\frac{1}{u(t)}$$ $$\frac{dv}{dt} =- t^3 - \frac{1}{t}v(t)$$ $$\frac{dv}{dt}+\frac{1}{t}v(t) =- t^3 \quad\text{is a linear ODE}.$$
I found out where that was my mistake: that's when I expanded $$t^3(y-t)^2$$ at the beggining.
$$\frac{dy}{dt}=t^3(y−t)^2+yt$$ $$\frac{dy}{dt}=t^3(y^2−2yt+t^2)+yt$$ $$\frac{dy}{dt}=t^3y^2−2yt^4+t^5+yt$$ $$...$$ So in the end of manipulation, I've got the same answer that could be found using @LutzL suggestion (replacing $$\frac1v = y-t$$): $$\frac{t^3}{v^2}+\frac{1}{vt}=\frac{-dv}{v^2 dt}$$ $$t^3+\frac{v}{t}=\frac{-dv}{dt}$$ Which is a linear ODE. I was really just missing a $$t^5$$ term. Sorry!