# Please check if the differential equation is correct

1. Feb 5, 2012

### rohit99

I am practising for my test. The question is to solve a differential equation

dy/dx + y/x + 1 = 5x

y(0) = 1.

The answer that i have come up with is

(xy+y)= 5x^3/3+5x^2/2+c

by substituting the values x=0 and y=1 in to the general equation I get

y(x+1)=5x^3/3 + 5x^2/2 +1

as the particular solution.

Can you tell me how will the particular solution look like and why this particular solution exists?

2. Feb 5, 2012

### chiro

Hey rohit99 and welcome to the forums.

For your equation your initial condition only effects a constant in the entire equation.

Based on this what is the effect of adding or subtracting a constant in a general equation? (Hint: how does it 'shift' the function?)

3. Feb 5, 2012

### HallsofIvy

Staff Emeritus
You mean y/(x+1).

What exactly do you mean by "look like"? If you just mean "solve for y", divide both sides by x+ 1.
The differential equation can be written
$$\frac{dy}{dx}= 5x- \frac{y}{x+1}$$
The function on the right side is differentiable for all y and all x except -1 so by the "fundamental existence and uniqueness theorem" for initial value problems, a unique solution to this problem exist for all x larger than -1.

Last edited: Feb 6, 2012