Particular Solution

Question

Let dy/dx = (y+1)/(x +1)  and let y = 3 when x = 1. The particular solution to this differential equation is y = 

A x.

B 2x.

C x + 2

D 2x + 1.

Answer 1

Hi,

Correct answer is D

dy/dx = (y+1)/(x +1)  and let y = 3 when x = 1

dy/dx = (y+1)/(x +1) both sides divide (y+1) and both sides time dx

then you have:

dy/(y+1) = dx/(x+1) now take integral of both sides

Ln(y+1) = Ln(x+1) + c for finding c plug in y = 3 when x = 1

Ln(3+1) = Ln(1+1) + c =====> c = Ln2 then

Ln(y+1) = Ln(x+1) + Ln2 use the formula such as Lna + Lnb = Lnab

Ln(y+1) = Ln(2x+2) =====> y+1 = 2x + 2 =====> y = 2x + 1

I hope it is useful,

Minoo

Answer 2

(1/(y+1))dy = (1/(x+1))dx

Integrate to get ln(y+1) = ln(x+1) + C

ln(y+1) - ln(x+1) = C

ln[(y+1)/(x+1)] = C

(y+1)/(x+1) = e^C

Relabel e^C as C to get y+1 = C(x+1)

Since y = 3 when x = 1, we have 4 = 2C. So, C = 2

y + 1 = 2(x+1)

Therefore, y = 2x + 1