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Please check if the differential equation is correct

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rohit99
#1
Feb5-12, 05:35 AM
P: 1
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?
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chiro
#2
Feb5-12, 06:53 AM
P: 4,578
Quote Quote by rohit99 View Post
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?
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?)
HallsofIvy
#3
Feb5-12, 04:47 PM
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P: 39,682
Quote Quote by rohit99 View Post
I am practising for my test. The question is to solve a differential equation

dy/dx + y/x + 1 = 5x
You mean y/(x+1).

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?
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
[tex]\frac{dy}{dx}= 5x- \frac{y}{x+1}[/tex]
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.


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