Question #22475

Show that the following equation has a solution of the form u(x,y) = e^(ax+by) and find the constants a and b:

U_xx + U_yy = 5e^(x-2y).

Expert's answer

Show that the following equation has a solution of the form u(x,y)=e(ax+by)u(x,y) = e^{(ax + by)} and find the constants aa and bb :


uxx+uyy=5e(x2y)u _ {x x} + u _ {y y} = 5 e ^ {(x - 2 y)}


**Solution:**


ux=ux=ae(ax+by)u _ {x} = \frac {\partial u}{\partial x} = a e ^ {(a x + b y)}uxx=2ux2=a2e(ax+by)u _ {x x} = \frac {\partial^ {2} u}{\partial x ^ {2}} = a ^ {2} e ^ {(a x + b y)}uy=uy=be(ax+by)u _ {y} = \frac {\partial u}{\partial y} = b e ^ {(a x + b y)}uyy=2uy2=b2e(ax+by)u _ {y y} = \frac {\partial^ {2} u}{\partial y ^ {2}} = b ^ {2} e ^ {(a x + b y)}uxx+uyy=a2e(ax+by)+b2e(ax+by)=(a2+b2)e(ax+by)u _ {x x} + u _ {y y} = a ^ {2} e ^ {(a x + b y)} + b ^ {2} e ^ {(a x + b y)} = (a ^ {2} + b ^ {2}) e ^ {(a x + b y)}(a2+b2)e(ax+by)=5e(x2y)a=1,b=2(a ^ {2} + b ^ {2}) e ^ {(a x + b y)} = 5 e ^ {(x - 2 y)} \rightarrow a = 1, b = - 2


Answer: a=1,b=2a = 1, b = -2 .

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