# Thread: Want help in solving this Diferrential equation by variable separable method!

1. ## Want help in solving this Diferrential equation by variable separable method!

How can I solve this euqation by variable separable ? I know after separating variable it will be solved by partial fraction by I stuck at the values of A and B on the side of y!

dy/dx = (y3 + 2y) / (x2 ​+ 3x )

2. ## Re: Want help in solving this Diferrential equation by variable separable method!

$\dfrac{1}{y^3+2y}=\dfrac{A}{y}+\dfrac{By+C}{y^2+2 }$

The numerator is always a polynomial of degree no greater than one less than the degree of the denominator.

$A=\dfrac{1}{2},B=-\dfrac{1}{2},C=0$

3. ## Re: Want help in solving this Diferrential equation by variable separable method!

May i know the steps of solution ?

4. ## Re: Want help in solving this Diferrential equation by variable separable method!

Originally Posted by damna97
May i know the steps of solution ?
It is one step.
$A (y^2+2)+(By+C)y=1$

$2A=1$ (coefficient of $y^0$)
$C=0$ (coefficient of $y^1$)
$A+B=0$ (coefficient of $y^2$)

5. ## Re: Want help in solving this Diferrential equation by variable separable method!

Step 1: take a Calculus class and learn to integrate!

You say yourself this equation is "separable"- so separate it. $\displaystyle \frac{dy}{dx}= \frac{y^3+ 2y}{x^2+ 3x}$ becomes $\displaystyle \frac{dy}{y^3+ 2y}= \frac{dx}{x^2+ 3x}$.

$\displaystyle \frac{dy}{y(y^2+ 2)}= \frac{dx}{x(x+ 3)}$

Now use "partial fractions" as you learned in Calculus to integrate those. That is what SlipEternal did.
$\displaystyle \frac{1}{y(y^2+ 2)}= \frac{A}{y}+ \frac{By+ C}{y^2+ 2}$
Multiply both sides by $\displaystyle y(y^2+ 2)$ to get $\displaystyle 1= A(y^2+ 2)+ (By+ C)y$. You need three equations to determine the three unknowns, A, B, and C, so substitute three different values for y to get them. If y= 0, 1= 2A. If y= 1, 1= 3A+ B+ C. If y= -1, 1= 3A+ B- C.

From 1= 2A, A= 1/2. The other two equations become 1= (3/2)+ B+ C or B+ C= 1- 3/2= -1/2 and 1= (3/2)+ B- C or B- C= 1- 3/2= -1/2. Adding those two equations 2B= -1 so B= -1/2. Subtracting 2C= 0 so C= 0 as slip eternal said.
$\displaystyle \frac{1}{y(y^2+ 2)}= \frac{1/2}{y}- \frac{1/2}{y^2+ 2}$.

On the other side you have $\displaystyle \frac{1}{x(x+ 3)}= \frac{A}{x}+ \frac{B}{x+ 3}$. (Of course, these are different "A" and "B" than before.)
Multiply on both sides by $\displaystyle x(x+ 3)$ to get $\displaystyle 1= A(x+ 3)+ Bx$. If x= 0, that becomes $\displaystyle 1= 3A$ so $\displaystyle A= \frac{1}{3}$. If x= -3, that becomes $\displaystyle 1= -3B$ so $\displaystyle B= -\frac{1}{3}$
$\displaystyle \frac{1}{x(x+ 3)}= \frac{1/3}{x}- \frac{1/3}{x+ 3}$.

So your equation finally reduces to
$\displaystyle \frac{1}{2}\frac{dy}{y}- \frac{1}{2}\frac{dy}{y^2+ 2}= \frac{1}{3}\frac{dx}{x}- \frac{1}{3}\frac{dx}{x+ 3}$.

Can you integrate that?

6. ## Re: Want help in solving this Diferrential equation by variable separable method!

Thankyou! Now I can solve this easily ! ☺️

7. ## Re: Want help in solving this Diferrential equation by variable separable method!

Originally Posted by HallsofIvy
Step 1: take a Calculus class and learn to integrate!

You say yourself this equation is "separable"- so separate it. $\displaystyle \frac{dy}{dx}= \frac{y^3+ 2y}{x^2+ 3x}$ becomes $\displaystyle \frac{dy}{y^3+ 2y}= \frac{dx}{x^2+ 3x}$.

$\displaystyle \frac{dy}{y(y^2+ 2)}= \frac{dx}{x(x+ 3)}$

Now use "partial fractions" as you learned in Calculus to integrate those. That is what SlipEternal did.
$\displaystyle \frac{1}{y(y^2+ 2)}= \frac{A}{y}+ \frac{By+ C}{y^2+ 2}$
Multiply both sides by $\displaystyle y(y^2+ 2)$ to get $\displaystyle 1= A(y^2+ 2)+ (By+ C)y$. You need three equations to determine the three unknowns, A, B, and C, so substitute three different values for y to get them. If y= 0, 1= 2A. If y= 1, 1= 3A+ B+ C. If y= -1, 1= 3A+ B- C.

From 1= 2A, A= 1/2. The other two equations become 1= (3/2)+ B+ C or B+ C= 1- 3/2= -1/2 and 1= (3/2)+ B- C or B- C= 1- 3/2= -1/2. Adding those two equations 2B= -1 so B= -1/2. Subtracting 2C= 0 so C= 0 as slip eternal said.
$\displaystyle \frac{1}{y(y^2+ 2)}= \frac{1/2}{y}- \frac{1/2}{y^2+ 2}$.

On the other side you have $\displaystyle \frac{1}{x(x+ 3)}= \frac{A}{x}+ \frac{B}{x+ 3}$. (Of course, these are different "A" and "B" than before.)
Multiply on both sides by $\displaystyle x(x+ 3)$ to get $\displaystyle 1= A(x+ 3)+ Bx$. If x= 0, that becomes $\displaystyle 1= 3A$ so $\displaystyle A= \frac{1}{3}$. If x= -3, that becomes $\displaystyle 1= -3B$ so $\displaystyle B= -\frac{1}{3}$
$\displaystyle \frac{1}{x(x+ 3)}= \frac{1/3}{x}- \frac{1/3}{x+ 3}$.

So your equation finally reduces to
$\displaystyle \frac{1}{2}\frac{dy}{y}- \frac{1}{2}\frac{dy}{y^2+ 2}= \frac{1}{3}\frac{dx}{x}- \frac{1}{3}\frac{dx}{x+ 3}$.

Can you integrate that?
Slight typo. Halls wrote that $B = -\dfrac{1}{2}, C=0$, then assigned $B=0, C=-\dfrac{1}{2}$. It should be:

$\displaystyle \dfrac{1}{y(y^2+2)} = \dfrac{1/2}{y} - \dfrac{y/2}{y^2+2}$

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$\displaystyle \dfrac{1}{2}\left( \dfrac{dy}{y} - \dfrac{y}{y^2+2}\right) = \dfrac{1}{3}\left(\dfrac{dx}{x} - \dfrac{dx}{x+3} \right)$

So, the LHS is actually easier than what Halls wrote (it will not require an inverse trigonometric integration).

Thanks.