non homogeneous differential equation

Question

I'm Leonardo and I'm a maths student from Trento,Italy.

I'm trying to solve analitically a Cauchy Problem which involves an ordonary differential equation of the form:

$\frac{dX(t)}{dt}+ a(t)X(t)^2=b(t)$ , with the following initial condition $X(0)=0$.

I know how to solve the corresponding homogeneous problem, with the method of separation of variables, but I don't know how to relate the solution of the homogeneous case, with the general problem. Is there any operative method?

I also know that the function $a(t)\rightarrow 0$ as $t\rightarrow 0$ and that the function $b(t)$ is linear.

Could anybody help me with this or give me some hints, please? That would be really appreciated :). Thanks to all for your time in advance.

Regards.

Answer 1

$$\frac{dX(t)}{dt}+ a(t)X(t)^2=b(t)$$ This is not a Bernoulli ODE : The Bernoulli ODEs are on the form : $\frac{dX(t)}{dt}+ a(t)X(t)^p=b(t)X(t)$. The term $b(t)X(t)$ instead of $b(t)$ makes a big difference.

The equation is a Riccati ODE.

Let : $X(t)=-\frac{y'(t)}{a(t)y(t)}\quad\to\quad X' =-\frac{y''}{ay} +\frac{(y')^2}{ay^2} +\frac{a'y'}{a^2y}$

$$-\frac{y''}{ay} +\frac{a'y'}{a^2y}=b$$

$$\frac{d^2y}{dt^2} -\frac{a'(t)}{a(t)} \frac{dy}{dt} +a(t) b(t) y(t)=0$$

Since $a(t)$ and $b(t)$ are not specified, the ODE is the general linear ODE of second order. We know that it can be solved for a solution on closed form only in particular casses, often involving special functions.

Thus, only in some cases depending on the kind of functions $a(t)$ and $b(t)$, an explicit form of solution $y(t)= c_1y_1(t)+c_2y_2(t)$ can be expressed. In thoses cases we derive : $$X(t)=-\frac{c_1y'_1(t)+c_2y'_2(t) }{a(t)\left(c_1y_1(t)+c_2y_2(t)\right) }$$

The condition $X(0)=0$ leads to : $$X(t)=-\frac{y'_2(0) y'_1(t)+y'_1(0)y'_2(t) }{a(t)\left(y'_2(0)y_1(t)+y'_1(0)y_2(t)\right) }$$

So, one cannot answer to the question about the condition $X(0)=0$ on a general manner. To go further, one have to know what kind of functions $a(t)$ and $b(t)$ are.