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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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1 Answer 1

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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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  • $\begingroup$ thank you! In the first equation of (2.): Where did the term "$+\frac{1}{C}y$" go? $\endgroup$
    – NW Patrick
    Commented Jun 24, 2011 at 16:00
  • $\begingroup$ More insights found here: voofie.com/content/6/… $\endgroup$
    – NW Patrick
    Commented Jun 25, 2011 at 19:44

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