# Differential Equation

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#1
Hi guys quick question,
Trying to solve this differential equation:

(x + 1)dy/dx - 3y = (x + 1)^5

Is the way to solve this one by expanding the bracket to the fifth power then splitting the variables? If it is, it seems a bit tedious but I can't think of another way of doing it. Any ideas?
0
5 years ago
#2
(Original post by Physics4Life)
Hi guys quick question,
Trying to solve this differential equation:

(x + 1)dy/dx - 3y = (x + 1)^5

Is the way to solve this one by expanding the bracket to the fifth power then splitting the variables? If it is, it seems a bit tedious but I can't think of another way of doing it. Any ideas?
Hi there,
The first thing you need to do is to make sure that the dy/dx stands on its own. This means getting rid of the (x+1) on the LHS by dividing the whole equation by (x+1).

It should leave you with:

dy/dx - 3y/(x+1)= (x+1)^4

Then you can work out the integrating factor from there and you should be able to work through to solve the equation!

Hope this helped, and if you need any more help just ask!

Vinstagram
0
#3
(Original post by Vinstagram)
Hi there,
The first thing you need to do is to make sure that the dy/dx stands on its own. This means getting rid of the (x+1) on the LHS by dividing the whole equation by (x+1).

It should leave you with:

dy/dx - 3y/(x+1)= (x+1)^4

Then you can work out the integrating factor from there and you should be able to work through to solve the equation!

Hope this helped, and if you need any more help just ask!

Vinstagram
Yes it did help thanks, also another question:
How did you recognise that this was an integrating factor question?
And also does dy/dx always have to be by itself when solving differential equations in general?
0
5 years ago
#4
(Original post by Physics4Life)
Yes it did help thanks, also another question:
How did you recognise that this was an integrating factor question?
And also does dy/dx always have to be by itself when solving differential equations in general?
I knew this simply because it did not have a d^2y/dx^2 term in the equation. A differential equation with a d^2y/dx^2 term is known as a 2nd order differential equation, whereas the one you mentioned above is a 1st order differential equation (generally consists of a dy/dx term + a something x y term on the left hand side of the '=' sign).

dy/dx always has to be by itself when solving a first order differential equation.
1
#5
(Original post by Vinstagram)
I knew this simply because it did not have a d^2y/dx^2 term in the equation. A differential equation with a d^2y/dx^2 term is known as a 2nd order differential equation, whereas the one you mentioned above is a 1st order differential equation (generally consists of a dy/dx term + a something x y term on the left hand side of the '=' sign).

dy/dx always has to be by itself when solving a first order differential equation.
0
5 years ago
#6
(Original post by Physics4Life)
No problem!
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