Question

a) Verify that the total differential equation
yz dx +(x^2y-zx)dy+(x^2z-xy)dz=0 is integrable and hence find its integral.

Answer 1

The given differential equation is ,

$yzdx+(x^2y-zx)dy+(x^2z-xy)dz=0$

We known that ,the necessary and sufficient condition of integrability of the given differential equation $Pdx+Qdy+Rdz=0$ is

$$P(\frac{\partial Q}{\partial z}-\frac{\partial R}{\partial y})+Q(\frac{\partial R}{\partial x}-\frac{\partial P}{\partial z} )+R(\frac{\partial P}{\partial y}-\frac{\partial Q}{\partial x})=0$$

$...........(1)$

Now ,according to question,

$P=yz,Q=(x^2y-zx) \ and \ R=(x^2z-xy)$

Now ,

$$\frac{\partial P}{\partial y}=z \ and \ \frac{\partial P}{\partial z}=y.$$ $$\frac{\partial Q}{\partial x}=(2xy-z) \ and \ \frac{\partial Q}{\partial z}=-x$$ $$\frac{\partial R}{\partial x}=(2xz-y) \ and \ \frac{\partial R }{\partial y}=-x$$

putting above all value in equation (1) ,we get ,

$yz(-x+x)+(x^2y-zx)\{(2xz-y)-(y) \}+(x^2z-xy) \{(z)-(2xy-z)\}$

$=0+2(x^3yz-x^2y^2-x^2z^2+xyz+x^2z^2-x^3yz-xyz+x^2y^2)=0$

Hence ,given differential equation is integrable.

To solve the given differential equation , we take

$x=constant. \ \implies \ dx=0$ .

$\therefore \ (x^2y-zx) dy+(x^2z-xy)dz=0$

$\implies \ x^2(ydy+zdz)-x(zdy+ydz)=0$

$\implies \ x^2(ydy+zdz)-xd(yz)=0$

On integrating ,we get

$$\frac{x^2 y^2} {2}+\frac{x^2z^2}{2}-xyz=\phi \ ........(2)$$

Where $\phi$ is a constant, it may be regarded as a function of $x.$

Differentiating (2) w.r.t. x , y, and z we get,

$(xy^2dx+yx^2dy)+(xz^2dx+zx^2dz)-(yzdx+xzdy+xydz)=d\phi$

On simplifying, we get

$(yx^2-xz)dy+(zx^2-xy)dz+yzdx+(xy^2+xz^2-2yz)dx=d\phi \ ...........(3)$

Since ,$yzdx+(yx^2-xz)dy+(zx^2-xy)dz=0 \ (given \ equation)$ ,therefore equation (3) is,

$(xy^2+xz^2-2yz)dx=d\phi \ ........(4)$

Again from equation (2),

$$\frac{x^2y^2}{2}+\frac{x^2z^2}{2}-xyz=\phi$$

$\implies \ x^2y^2+x^2z^2-2xyz=2\phi$

Now dividing both side by x ,we get

$$\implies \ xy^2+xz^2-2yz=\frac{2\phi}{x}............(5)$$

From equation (4) and (5) ,we get

$$\frac{2\phi }{x} dx=d\phi$$ $$\implies \ \frac{2dx}{x}=\frac{d\phi }{\phi}$$

On integrating ,we get

$logx^2+logk=log\phi$

Where k is a constant.

$\implies \ log(x^2k)=log\phi$

$\implies \ x^2k=\phi$

Now ,putting the value of $\phi$ in equation (2) ,we get

$$\frac{x^2y^2}{2}+\frac{x^2z^2}{2}-xyz=x^2k$$

$\implies \ xy^2+xz^2-2yz=2xk$