Question

Which one of the following is the general solution of the first order differential equation

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

where x , y are real?

• A. y = 1 + x + $ta{n}^{-1}(x+c),$ where c is a constant
• B. y = 1 + x + $tan(x+c),$ where c is a constant
• C. y = 1 - x + $ta{n}^{-1}(x+c),$ where c is a constant
• D. y = 1 - x + $tan(x+c),$ where c is a constant

Answer 1

The correct option is D y = 1 - x + $tan(x+c),$ where c is a constant

Given,

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

Let x + y -1 = t

Then 1 + $\frac{dy}{dx}-0=\frac{dt}{dx}$

Or $\frac{dy}{dx}=\frac{dt}{dx}-1$

From equation (1)

$\frac{dt}{dx}-1=t^2$

$\frac{dt}{1+t^2}=dx$

$\int \frac{dt}{1+t^2}=\int dx$

$ta{n}^{-1}t=x+c$

t = tan (x + c)

x + y - 1 = tan (x + c)

y = 1 - x + tan (x + c)