Question

Let y = y 1(x) and y = y 2(x) be two distinct solution of the differential equation
$\frac{dy}{dx} = x+y,$
with y 1(0) = 0 and y 2(0) = 1 respectively. Then, the number of points of intersection of y = y 1 (x) and y = y 2(x) is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (A)

0

Answer 2

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 y 1(x) and y 2(x).