If Z - eax-byFax-by- the value of b- z- x-a- z- y is

Z = e^ax+byF(ax by) ∂ z∂ x=ae^ax+byF(ax by)+ e^ax+byaF'(ax by) and ∂ z∂ y=be^ax+byF(ax by)+e^ax by( b)F'(ax by) Now b∂ z∂ x+a∂ z∂ y=2abe^ax+yF(ax by) = 2ab ...

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Engineering Mathematics

Method of Variation of Parameters

If Z = eax+by...

Question

If Z = $e^{a x + b y} F (a x - b y)$ ; the value of $b \frac{\partial z}{\partial x} + a \frac{\partial z}{\partial y}$ is

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2abZ

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Solution

The correct option is D 2abZ
z = $e^{a x + b y} F (a x - b y)$

$\frac{\partial z}{\partial x} = a e^{a x + b y} F (a x - b y) + e^{a x + b y} a F^{'} (a x - b y)$

and $\frac{\partial z}{\partial y} = b e^{a x + b y} F (a x - b y) + e^{a x - b y} (- b) F^{'} (a x - b y)$

Now $b \frac{\partial z}{\partial x} + a \frac{\partial z}{\partial y} = 2 a b e^{a x + y} F (a x - b y)$

= 2ab.z

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If Z = eax+byF(ax−by); the value of b∂z∂x+a∂z∂y is

The correct option is D 2abZz = eax+byF(ax−by) ∂z∂x=aeax+byF(ax−by)+eax+byaF′(ax−by) and ∂z∂y=beax+byF(ax−by)+eax−by(−b)F′(ax−by) Now b∂z∂x+a∂z∂y=2abeax+yF(ax−by) = 2ab.z

If Z = $e^{a x + b y} F (a x - b y)$ ; the value of $b \frac{\partial z}{\partial x} + a \frac{\partial z}{\partial y}$ is