nonlinear, nonhomogeneous ODE 1. order

Question

Solve $x'(t)-\dfrac{a}{t}x(t)=b(t),~a=const,~x(0)=0$.

Homogeneous Solution:

$\dfrac{x'(t)}{x(t)}=\dfrac{a}{t}\quad|\int\\ \ln(x(t))=a\ln(t)+c,~c=const\quad| e\\ x(t)=t^ae^c$

Is that correct?

No clue how to figure the particular solution or the general solution respectively.

edit: Thanks @agha for the helpfull comment. I try to write down the entire solution. With a slighty different Noation.

Let $t\in[0,t_f]$ $$(1)~ x'(t) - \dfrac{a}{t}x(t) = b(t)$$ $$(2)~ x_h(t) = Kt^a$$

Particular Solution: $$x_s(t) = c(t)t^a~\text{with}~ x'(t) = c'(t)t^a + c(t)at^{a-1}$$

Substitute in (1) $$c'(t)t^a + c(t)at^{a-1} - \dfrac{a}{t}c(t)t^a = b(t)$$ $$c'(t) = \dfrac{b(t)}{t^a}\quad |\int{}$$ $$c(t) = \int{\dfrac{b(t)}{t^a}dt} + K'$$

It follows for $x_s(t)$ $$(3)~ x_s(t) = \left(\int{\dfrac{b(t)}{t^a}dt} + K'\right)t^a$$

The general solution reads (2)+(3), i.e. $$(4)~ x(t) = Kt^a + \left(\int{\dfrac{b(t)}{t^a}dt} + K'\right)t^a = \left(\int{\dfrac{b(t)}{t^a}dt} + K''\right)t^a\\ \text{where}~ K''=K+K'$$

With the initial conditon $x(0)=0$ we might identify $K''$ $$(5)~ x(0) = 0 = \int{\dfrac{b(t)}{t^a}dt}\bigg|_{t_f} + K''\\ K'' = -\int{\dfrac{b(t)}{t^a}dt}\bigg|_{t_f}$$

Thus the solution reads: $$(6)~x(t) = \left(\int{\dfrac{b(t)}{t^a}dt} - \int{\dfrac{b(t)}{t^a}dt}\bigg|_{t_f}\right)t^a$$

Am I right?

Answer 1

You can use general method - Variation of parameters.

Let $e^C=D$, so you have solution of homogeneous ODE. $x(t)=Dt^{a}$ ($D$ is a constant). Now by variation of parameter solution of non-homogeneous ODE is $x(t)=D(t)t^{a}$($D$ is a function).Next:

$$x'(t)=(D(t)t^{a})'=D'(t)t^{a}+aD(t)t^{a-1}$$

Now substitute to equation: $$x'(t)-\frac{a}{t}x(t)=(D'(t)t^{a}+aD(t)t^{a-1})-\frac{a}{t}D(t)t^{a}=D'(t)t^{a}=b(t)$$

So:

$$D'(t)=\frac{b(t)}{t^{a}}$$

Finally:

$$D(t)=\int \frac{b(t)}{t^{a}} dt $$