Solve the ODE $\frac{\mathrm dx}{\mathrm dt}=(x+t)t$

$\def\d{\mathrm{d}}$How to solve this ODE? (From a real analysis course, existence and uniqueness of ODE) $$\frac{\d x}{\d t}=(x+t)t. \quad \forall t\in [0,1], \quad x(0)=0$$

My attempt:

$$\dot{x}=\frac{\d x}{\d t}=V(x(t),t)=(x+t)t$$

So we can use $\phi_v(x)$ such that

$$\phi_v^1(x,t)=x(0)+\int_{0}^{t}(x+t)t\,\d t= \frac{t^3}{3}+\frac{t^2x}{2}$$ with $x(0)=0$.

Applying $\phi_v$ to approximate $\phi^n_v$:

$$\phi_v^2(x,t)=\int_{0}^{t} \left( \frac{t^3x}{2}+\frac{t^4}{3}+t^2 \right) \, \d t= \frac{t^4x}{8}+\frac{t^5}{15}+\frac{t^3}{3},\\ \phi_v^3(x,t)=\int_0^t \left( \frac{t^5x}{8}+\frac{t^5}{15}+t^4 \right) \,\d t= \frac{t^6x}{48} +\frac{t^7}{106}+\frac{t^3}{15}+\frac{t^3}{3}.$$

Am I on the right track? How should I finish this?

• The solution to $x'=(x+t)t$ ends up using the Erf function. To help people tailor an answer for you, can you provide some background for the question - where did it come from?, what level of calculus are you at? – Ian Miller Apr 20 '17 at 15:35
• @IanMiller this question is from a real analysis course. Topic is existance and uniqueness of ODE – combo student Apr 20 '17 at 15:37
• @IanMiller How do i get to that? – combo student Apr 20 '17 at 15:42

HINT

Substitute $y(t) = x(t)+t$ so $\dot{y} = \dot{x} +1$ and your ODE becomes the familiar linear $$\frac{dy}{dt} = 1 + yt$$ with initial condition $y(0) = x(0)+0 = 0$.

• Isn't $\dot x-xt=t^2$ as familiar ? – Yves Daoust Apr 20 '17 at 15:56

Let $u=x+t$ ,

Then $x=u-t$

$\dfrac{dx}{dt}=\dfrac{du}{dt}-1$

$\therefore\dfrac{du}{dt}-1=tu$

$\dfrac{du}{dt}-tu=1$

I.F. $=e^{-\int t~dt}=e^{-\frac{t^2}{2}}$

$\therefore\dfrac{d\left(e^{-\frac{t^2}{2}}u\right)}{dt}=e^{-\frac{t^2}{2}}$

$e^{-\frac{t^2}{2}}u=\int_0^te^{-\frac{\tau^2}{2}}~d\tau$

$x+t=e^\frac{t^2}{2}\int_0^te^{-\frac{\tau^2}{2}}~d\tau$

$x=e^\frac{t^2}{2}\int_0^te^{-\frac{\tau^2}{2}}~d\tau-t$