Question

Let y = y₁(x) and y = y₂(x) be two distinct solution of the differential equation
\(\frac{dy}{dx} = x+y,\)
with y₁(0) = 0 and y₂(0) = 1 respectively. Then, the number of points of intersection of y = y₁ (x) and y = y₂(x) is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The Correct Option is A

The correct answer \(\frac{dy}{dx}\) is (A):
\(\frac{dy}{dx} = x+y\)
Let x + y = t
1+\(\frac{dy}{dx}\)=\(\frac{dt}{dx}\)
\(\frac{dt}{dx}\)-1 = t
⇒ ∫\(\frac{dt}{t}\)+1 = ∫dx
In |t+1| = x+C'
|t+1|=Cex
|x+y+1| = Cex
For y1 (x),y1(0) = 0
⇒ C = 1
For y2 (x),y2(0) = 1
⇒ C = 2
y1(x) is given by |x+y+1| = ex
y2(x) is given by |x+y+1| = 2ex
At point of intersection
ex = 2ex
No solution
So, there is no point of intersection of y₁(x) and y₂(x).