Question

Eliminate the arbitrary constants and form corresponding partial differential equation Z=ax^2 + bxy + cy^2​

Answer 1

Partial Differential Equation

Given: $\mathrm{z=ax^{2}+by^{2}+cy^{2}}$

To find:

• eliminate the arbitrary constants
• form corresponding partial differential equation

Solution:

Given $\mathrm{z=ax^{2}+bxy+cy^{2}}$ .....(1)

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• Differentiating $\mathrm{z}$ with respect to $\mathrm{x}$, we get

• $\mathrm{\frac{\partial z}{\partial x}=2ax+by}$

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• Differentiating $\mathrm{z}$ with respect to $\mathrm{y}$, we get

• $\mathrm{\frac{\partial z}{\partial y}=bx+2cy}$

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• Again differentiating $\mathrm{\frac{\partial z}{\partial x}}$ with respect to $\mathrm{x}$, we get

• $\mathrm{\frac{\partial ^{2}z}{\partial x^{2}}=2a}$

• $\mathrm{\Rightarrow a=\frac{1}{2}\frac{\partial ^{2}z}{\partial x^{2}}}$

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• Now differentiating $\mathrm{\frac{\partial z}{\partial x}}$ with respect to $\mathrm{y}$ and same for $\mathrm{\frac{\partial z}{\partial y}}$ with respect to $\mathrm{x}$, in both cases, we get

• $\mathrm{\frac{\partial ^{2}z}{\partial x\partial y}=\frac{\partial ^{2}z}{\partial y\partial x}=b}$

• $\mathrm{\Rightarrow b=\frac{\partial ^{2}z}{\partial x\partial y}}$

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• Also differentiating $\mathrm{\frac{\partial z}{\partial y}}$ with respect to $\mathrm{y}$, we get

• $\mathrm{\frac{\partial ^{2}z}{\partial y^{2}}=2c}$

• $\Rightarrow \mathrm{c=\frac{1}{2}\frac{\partial ^{2}z}{\partial y^{2}}}$

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We have found:

• $\mathrm{a=\frac{1}{2}\frac{\partial ^{2}z}{\partial x^{2}}}$

• $\mathrm{b=\frac{\partial ^{2}z}{\partial x\partial y}}$

• $\mathrm{c=\frac{1}{2}\frac{\partial ^{2}z}{\partial y^{2}}}$

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Substituting the values of the arbitrary constants $\mathrm{a}$, $\mathrm{b}$ and $\mathrm{c}$ in (1), we get

• $\mathrm{z=\frac{1}{2}\frac{\partial ^{2}z}{\partial x^{2}}x^{2}+\frac{\partial ^{2}z}{\partial x\partial y}xy+\frac{1}{2}\frac{\partial ^{2}z}{\partial y^{2}}y^{2}}$

This is the required partial differential equation.