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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?

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    $\ln (P - K) = \lambda t + C$ should be $- \ln (P - K) = \lambda t + C$.2012-10-31

1 Answers 1

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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 $K(0) = 0.2P$.