First Order ODE - Variation Of Parameters

Question

$$\frac{dC}{dt}=a-bC\text{ , }C(0)=K$$ Where $a,b$ are positive constants

can use variation of parameters? or because we need ODE in the form of $C'+P(t)C=Q(t)$ but $a,b$ are not function of $t$, we can not use it?

If so can we use variation of parameters? on $y'-\frac{y}{x}=1$ or not because $1$ is not a function of $x$?

why not? the constant function is a function of $t$ as any other function — tired, Jan 18 '17 at 20:16
Furthermore, the solution inhomogenous to this equation can be simply read of: $C(t)=a/b$ — tired, Jan 18 '17 at 20:17
Ok got it, I can also bring it to an homogeneous by $\frac{dC}{dt}+b(-\frac{a}{b}+C)=0$ and solve by separation of variables — gbox, Jan 18 '17 at 22:01

score 3 · Accepted Answer · answered Jan 18 '17 at 20:29

You can use variation of parameters in both cases.

For the first equation, you have:

$$\frac{dC}{dt}+bC=a \tag{1}$$

The homogeneous form (when $a=0$ in this case) will be:

$$\frac{dC}{dt}=-bC \tag{2}$$

Which can be evaluated to give:

$$\ln|C|=-bt+k$$ $$C(t)=Ke^{-bt} \tag{3}$$

Equation $(3)$ is the solution to your homogeneous equation.

Now, just let $K$ vary in terms of $t$:

$$C(t)=K(t)e^{-bt} \tag{4}$$ And differentiate this: $$C'(t)=-b K(t)e^{-bt}+K'(t)e^{-bt}$$

Now, you can substitute this into your original ODE on $(1)$:

$$-b K(t)e^{-bt}+K'(t)e^{-bt}+bK(t)e^{-bt}=a$$

This should cancel out nicely, and you should be able to solve for $K(t)$, which can be substituted into $(4)$, giving the general solution to the full differential equation.

You can apply this same method to your other differential equation $\frac{dy}{dx}-\frac{y}{x}=1$ by letting $1$ equal $0$ to find a solution to your homogeneous equation.

First Order ODE - Variation Of Parameters

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