Question

obtain all the first and second order PD of the fuction f(x,y)=(x^2)Siny+(y^2)cosx

Answer 1

Answer on Question #75735 - Subject – Differential Equation

**Given:** $f(x, y) = x^{2} \sin y + y^{2} \cos x$

**To find:** All the first and second order partial differential equations

**Solution:** Consider $f(x, y) = x^{2} \sin y + y^{2} \cos x$

Differentiating partially with respect to $x$ and $y$, we get

$$\frac{\partial f}{\partial x} = 2x \sin y - y^{2} \sin x \quad \text{and} \quad \frac{\partial f}{\partial y} = x^{2} \cos y + 2y \cos x$$ $$\therefore x \frac{\partial f}{\partial x} = 2x^{2} \sin y - x y^{2} \sin x \quad \text{and} \quad y \frac{\partial f}{\partial y} = x^{2} y \cos y + 2y^{2} \cos x$$ $$\Rightarrow x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = 2x^{2} \sin y - x y^{2} \sin x + x^{2} y \cos y + 2y^{2} \cos x$$ $$\Rightarrow x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = 2(x^{2} \sin y + y^{2} \cos x) - x y (y \sin x - x \cos y)$$ $$\Rightarrow x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = 2f - x y (y \sin x - x \cos y)$$

Again differentiating,

$$\frac{\partial^{2} f}{\partial x^{2}} = 2 \sin y - y^{2} \cos x, \quad \frac{\partial^{2} f}{\partial y^{2}} = -x^{2} \sin y + 2 \cos x \quad \text{and} \quad \frac{\partial^{2} f}{\partial x \partial y} = 2 (x \cos y - y \sin x)$$

From equation (i)

$$y \sin x - x \cos y = \frac{1}{x y} \left(2f - x \frac{\partial f}{\partial x} - y \frac{\partial f}{\partial y}\right)$$ $$\Rightarrow x \cos y - y \sin x = \frac{1}{x y} \left(x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} - 2f\right)$$ $$\Rightarrow \frac{\partial^{2} f}{\partial x^{2}} = \frac{1}{x y} \left(x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} - 2f\right)$$

using second order derivative.

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