# How to solve this differential equation?

I want to solve this differential equation:

$$C \cdot y(t)\frac{d}{dt} = x(t) - y(t)$$

$x(t)$ and $y(t)$ are two ordinary functions of t, C is a constant - all in in $R$

I am trying to solve it towards $y(t)$. The solution I am looking for looks something like this:

$$y(t) = e^{\int{x(t)dt}} + C$$

So $\int{x(t)dt}$ can stay - but how will the rest look like and could you show me the individual steps and name the method how to solve it?

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This comes very close to what I'm trying to do but my equation is a little bit simpler: de.wikipedia.org/wiki/Variation_der_Konstanten –  NW Patrick Jun 24 '11 at 14:27
What do you differentiate in your equation? The differentiation operator is written before the function, not after it. stewartcalculus.com/data/CALCULUS%20Concepts%20and%20Contexts/… is a nice document you should take a look at. –  Beni Bogosel Jun 24 '11 at 14:31

If you mean an equation $Cy'(t) = x(t)-y(t)$ then the solution is following.

If $C=0$ then $y = x$. If $C\neq 0$ then $y' + \frac 1 Cy = \frac{x}{C}$. You can solve it by the method "Variation of constant" or "Lagrange method".

1. Solve the homogeneous equation: $$y'+\frac{1}{C}y = 0.$$ Separate variables: $$\frac{dy}{y} = -\frac 1 C dt.$$ By integration we obtain $$y=K\exp\{-t/C\}$$ where $K$ is some constant.

2. In Lagrange method you suppose that $K(t)$ is a function rather than a constant and substitute $y=K(t)\exp\{-t/C\}$ in the original equation $y'+\frac 1 C y = \frac x C$ to find $K(t).$ After substituition you obtain: $$K'\exp\{-t/C\} = \frac{x}{C},$$ so $$K' = \frac{1}{C}x(t)\exp\{t/C\}$$ and $$K = K_1+\frac{1}{C}\int x(t)\exp\{t/C\}dt.$$

Now, $$y(t) = \left(K_1+\frac{1}{C}\int x(t)\exp\{t/C\}dt\right)\exp\{-t/C\}$$ where $K_1$ is some constant.

By the way, $\exp$ means exponent in this notation, $\exp\{a\} =\mathrm e^a$.

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thank you! In the first equation of (2.): Where did the term "$+\frac{1}{C}y$" go? –  NW Patrick Jun 24 '11 at 16:00
More insights found here: voofie.com/content/6/… –  NW Patrick Jun 25 '11 at 19:44