Question

Solve following differential equations (x-y)^2 dy/dx=a^2.

Answer 1

solution

(x-y)²dy/dx=a²

2(x-y)(1-dy/dx)=0

(2x-2y)(1-dy/dx)=0

2x-2dy/dx-2y+2ydy/dx=0

-2dy/dx+2ydy/dx=2y-2x

⇒ dy/dx=(2y-2x)/(2y-2)

Answer 2

Answer:

The solution for the given differential equation $(x-y)^2\frac{dy}{dx}=a^2$  is $a\times\ \tan^{-1}(x-y)=y+c$.

Step-by-step explanation:

Consider, the given differential equation,

$(x-y)^2\frac{dy}{dx}=a^2$   ..............(1)

Put $x-y=u$ then differentiate with respect to x, we get

$1-\frac{dy}{dx}=\frac{du}{dx}$

$\frac{dy}{dx}=1-\frac{du}{dx}$

Substitute in (1), then $(x-y)^2\frac{dy}{dx}=a^2$ becomes,

$u^2(1-\frac{du}{dx})=a^2$

$u^2-u^2\frac{du}{dx})=a^2$

$u^2\frac{du}{dx})=u^2-a^2$

Taking similar terms together,

$\frac{u^2}{u^2-a^2} {du}=dx$

Add and Subtract $a^2$ in left side, then

$\frac{u^2-a^2+a^2}{u^2-a^2} {du}=dx$

Splitting terms,

$(\frac{u^2-a^2}{u^2-a^2}+\frac{a^2}{u^2-a^2}){du}=dx$

$(1+\frac{a^2}{u^2-a^2}){du}=dx$

Differentiate both sides,

$\int{du}+\int\frac{a^2}{u^2-a^2}{du}=\int{dx}$

$\int(\frac{1}{u^2-a^2})du=\frac{1}{a} \tan^{-1}u$

$u+a^2\times\frac{1}{a} \tan^{-1}u=x+c$

$u+a\times\ \tan^{-1}u=x+c$

Put value of u = x-y

$(x-y)+a\times\ \tan^{-1}(x-y)=x+c$

$-y+a\times\ \tan^{-1}(x-y)=c$

$a\times\ \tan^{-1}(x-y)=y+c$

Thus, the solution for the given differential equation $(x-y)^2\frac{dy}{dx}=a^2$  is $a\times\ \tan^{-1}(x-y)=y+c$.