2
$\begingroup$

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?

  • 0
    What do you differentiate in your equation? The differentiation operator is written before the function, not after it. http://www.stewartcalculus.com/data/CALCULUS%20Concepts%20and%20Contexts/upfiles/3c3-LinearDiffEqns_Stu.pdf is a nice document you should take a look at.2011-06-24

1 Answers 1

3

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, $\exp\{a\} =\mathrm e^a$.

  • 0
    More insights found here: http://www.voofie.com/content/6/introduction-to-differential-equation-and-solving-linear-differential-equations-using-operator-metho/#First_order_Linear_Differential_Equations2011-06-25