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If z = tan(y+ax) + (y-ax)(3/2)
\[\dfrac{\partial^2z}{\partial x^2} =a^2 \dfrac{\partial^2z}{\partial y^2}\]
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written 3.0 years ago by
teamques10
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$\dfrac {\partial z}{\partial x} = a \sec^2 (y+ax) - \left ( \dfrac {3a}{2} \right ) (y-ax)^{1/2} $ $\text{and, } \dfrac {\partial^2 z}{\partial x^2} = a^2 2\sec^2 (y+ax)\tan (y+ax) + \left ( \dfrac {3a^2}{4} \right ) (y-ax)^{-1/2} \cdots \ \cdots (1)$ $ \text{now, } \dfrac {\partial z}{\partial y} = \sec^2 (y+ax) …
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