Question

(b) dy/dx = (x-y + 1)^2

Answer 1

Answer:

The required answer is $x+C=\frac{1}{2}\ln|\frac{2+x-y}{y-x}|$.

Step-by-step explanation:

The given differential equation is

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

Substitute u=x-y+1 in the above equation.

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

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

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

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

Using variable separable method, we get

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

Integrate both the sides.

$\int dx=\int \frac{du}{1-u^2}$

$x+C=\frac{1}{2}\ln|\frac{1+u}{1-u}|$      $[\because \int \frac{dx}{a^2-x^2}=\frac{1}{2a}\n|\frac{a+x}{a-x}|+C]$

Substitute u=x-y+1 in the above equation.

$x+C=\frac{1}{2}\ln|\frac{1+x-y+1}{1-(x-y+1)}|$

$x+C=\frac{1}{2}\ln|\frac{2+x-y}{y-x}|$

Therefore the required answer is $x+C=\frac{1}{2}\ln|\frac{2+x-y}{y-x}|$.