This question from my problem set has been driving me crazy:
A differential equation of the form $y' = P(x) + Q(x)y + R(x)y^2$
is called Riccati’s equation. In general, the equation cannot be solved by elementary methods. But if a particular solution $y = y_1(x)$ is known, then the general solution is given by $y(x) = y_1(x) + u(x)$ where $u$ satisfies the Bernoulli equation $$\frac{du}{dx} − (Q + 2Ry_1)u = Ru^2$$ Use the method to solve $y' + x^3y − x^2y^2 = 1$, given $y_1 = x$.
My approach: Rearranging, $y'=1-x^3y+x^2y^2$. So $Q=-x^3$ and $R=x^2$ we get $\frac{du}{dx}-x^3u=u^2x^2$
Now, converting to Bernoulli's form and taking $\frac{1}{u}=t$, the differential equation reduces to $\frac{dt}{dx}+x^3t=-x^2$
This is where the trouble begins. The integration factor comes out to be $e^\frac{x^4}{4}$. How do I calculate the value of $\int-x^2e^\frac{x^4}{4}$? Or have I made a silly blunder somewhere that has caused this?
Best Answer
$$u'-x^3u=u^2x^2$$ Substitute $v=\dfrac 1 u \implies v'=-\dfrac {u'}{u^2}$: $$v'+x^3v=-x^2$$ $$(ve^{\frac {x^4}4})'=-x^2e^{\frac {x^4}4}$$ $$\dfrac 1 {u(x)}=-e^{-\frac {x^4}4}\int x^2e^{\frac {x^4}4}dx+Ce^{-\frac {x^4}4}$$ The RHS can't be integrated with elementary functions. You made no mistakes. You can keep the integral in the solution.