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Can anyone help me see how one can get from the following integral

$$\lambda\int_{\beta=0}^{y}\int_{\alpha=x}^{Q-\beta}e^{-\mu(\alpha-x)}f(\alpha,\beta)d\alpha d\beta+k\int_{\alpha=s}^{x}f(\alpha,y)d\alpha$$ $$ = $$ $$\lambda\int_{\beta=0}^{y}\int_{\alpha=s}^{Q-\beta}e^{-\mu(\alpha-s)}f(\alpha,\beta)d\alpha d\beta+k\int_{\alpha=s}^{x}f(\alpha,0)d\alpha$$ to this second order partial differntial equation $\frac{\partial^{2}}{\partial x\partial y}f(x,y)-\mu\frac{\partial}{\partial y}f(x,y)-\frac{\lambda}{k}\frac{\partial}{\partial x}f(x,y)=0$, where $\mu,k,\lambda,Q,s$ are all constants.

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I would also suggest dropping all instances of "$\alpha=$" and "$\beta=$" from the equation. –  Ricky Demer Jun 18 '11 at 22:30
Have you tried just taking the $\partial^2/\partial x \partial y$ derivative and applying the fundamental theorem of calculus? –  Willie Wong Jun 19 '11 at 1:07
Yes I took $\frac{\partial}{\partial y}$ once and $\frac{\partial}{\partial x}$ twice from both sides but I couldn't get the above form... –  V.A. Jun 19 '11 at 1:16

1 Answer 1

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Let $U =$ $$ \lambda \int_{0}^{y} \int_{x}^{Q - \beta} \operatorname{e} ^{\bigl(-\mu (\alpha - x)\bigr)} f (\alpha,\beta) d \alpha d \beta - \lambda \int_{0}^{y} \int_{s}^{Q - \beta} \operatorname{e} ^{\bigl(-\mu (\alpha - s)\bigr)} f (\alpha,\beta) d \alpha d \beta + k \int_{s}^{x} f (\alpha,y) d \alpha - k \int_{s}^{x} f (\alpha,0) d \alpha $$ so that our equation is $U=0$. Then compute $$ 0 = \frac{\partial^{3} U}{\partial y \partial x^{2}} - \mu \frac{\partial^{2} U}{\partial y \partial x} = -\lambda \frac{\partial f (x,y)}{\partial x} + k \frac{\partial^{2} f (x,y)}{\partial y \partial x} - \mu k \frac{\partial f (x,y)}{\partial y} $$

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