# Autonomous linear ODE

Given $K(0) = 0,\!2P$. I'm supposed to solve the ODE

$$\frac{dK}{dt} = \lambda(P-K)$$

I have tried to seperate and integrate both sides

$$\int \frac{1}{P-K} dK = \int \lambda \space dt$$

to get

$$\ln(P-K) = \lambda t + C$$

and then solve for $K$

$$e^{\ln(P-K)} = P-K=e^{\lambda t + C}$$ $$-(-K) = -(-P + e^{\lambda t + C})$$ $$K = P - e^{\lambda t + C}$$

is that about right for the general solution? And where I'm really stuck is how do I proceed to find the particular solution?

-
$\ln (P - K) = \lambda t + C$ should be $- \ln (P - K) = \lambda t + C$. –  glebovg Oct 31 '12 at 0:14

Note that $\ln (P - K) = \lambda t + C$ should be $- \ln (P - K) = \lambda t + C$ because you need to use the substitution $u = P - K$ so that $du = - dK$. To find the particular solution use the initial data. Set $t = 0$ and $K = 0.2P$ to solve for $C$, since $K(0) = 0.2P$.