FIRST ORDER ODE QUESTION

[PapayaC]

So I did the question but i don´t know if my answer is right, can someone please help me?

Question: Find the general solution of the following ODE:

(x² + x)y‘ = 3y – 1

Answer: (x² + x)dy/dx = 3y – 1

(x^(2)+x)dx=1/(3y-1)dy
(x^3)/3+ (x^2)/2= ln(3y-1)*(1/3)
y= (e^(x^3+1.5x^2)+1)/3

Is this answer right? It looks so messy so I´m not sure. Any help would be really appreciated.

 

[topsquark]

So I did the question but i don´t know if my answer is right, can someone please help me?

Question: Find the general solution of the following ODE:

(x² + x)y‘ = 3y – 1

Answer: (x² + x)dy/dx = 3y – 1

(x^(2)+x)dx=1/(3y-1)dy

You didn't separate this right. The last line should read
[math]\dfrac{dx}{x^2 + x} = \dfrac{dy}{3y - 1}[/math]
-Dan

 

[Deleted member 4993]

So I did the question but i don´t know if my answer is right, can someone please help me?

Question: Find the general solution of the following ODE:

(x² + x)y‘ = 3y – 1

Answer: (x² + x)dy/dx = 3y – 1

(x^(2)+x)dx=1/(3y-1)dy
(x^3)/3+ (x^2)/2= ln(3y-1)*(1/3)
y= (e^(x^3+1.5x^2)+1)/3

Is this answer right? It looks so messy so I´m not sure. Any help would be really appreciated.

The last response pointed out that you did not "separate" the variables correctly. In addition, your answer did not include constant of integration.

I'll give you one more step beyond that:

(x² + x)y‘ = 3y – 1

dx/(x² + x) = dy/(3y -1)

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

Now integrate and don't forget the constant of integration.

 

[JeffM]

First, note that if you are going to try to separate variables, you need to pay attention to where various expressions are positive, negative, or zero unless you have been told, but did not tell us, that there are restrictions on the domain of the function y(x).

Second, it may not be obvious what Subhotosh did. (It took me a few minutes to see.)

[MATH]\dfrac{dy}{3y - 1} = \dfrac{dx}{x^2 + x} = \dfrac{dx}{x(x + 1)} = \dfrac{A \ dx}{x} + \dfrac{B \ dx}{x + 1}.[/MATH]
But what are A and B? Well, it must be true that

[MATH]\{A(x + 1) + Bx\} \ dx = dx \implies A(x + 1) + Bx = 1 \implies \\ Ax + A + Bx = 1 \implies x(A + B) + A = 1. \\ \text {But } x \text { is a VARIABLE. Therefore IF } A \text { and } B \\ \text { are constants, then } A + B \text { must } = 0 \implies A = 1 \implies B = -1.[/MATH]Of course we cannot be sure that A and B are constants. Let's check.

[MATH]\dfrac{1}{x} + \dfrac{-1}{x + 1} = \dfrac{(1)(x + 1)}{x(x + 1)} + \dfrac{(-1)x}{x(x + 1)} = \dfrac{x + 1 - x}{x^2 + x} = \dfrac{1}{x^2 + 1} \ \checkmark.[/MATH]

 

[Steven G]

Why not differentiate your answer to see if have the correct results?