General solution of differential equation?

Question

What is the general solution of (2+x)y'=3y? I've gotten to e^((1/3)lny) = e^(ln(2+x)+c) , but I don't know how to simplify it further.

Answer 1

(2 + x) y' = 3y ⇒ (2 + x) dy/dx = 3y ⇒ dy/3y = dx/(2 + x) ⇒ (1/3)ln(y) = ln(2 + x) + c

 

Here we can use the property (a)ln(x) = ln(x^a)

ln(y^(1/3)) = ln(2 + x) + c ⇒ e^{ln(y^(1/3))} = e^{(ln(2 + x) + c)} 

 

Now we use the property x^{(a + b)} = (x^a)(x^b)

y^(1/3) = e^{ln(2 + x)} * e^c ⇒ y^(1/3) = (e^c)(2 + x) ⇒ y = (e^c)³(2 + x)³

 

Since (e^c)³ is still a constant we can call it something else that is still constant like k.

y = k(2 + x)³

 

Lets check our answer.

y = k(2 + x)³ and y' = 3k(2 + x)²

 

So,

(2 + x)3k(2 + x)² = 3k(2 + x)³ 

As we can see they are both equal and our answer is correct.

Answer 2

1/3 ln y = ln (2 + x) + c.  Since it is a natural log equation the 1/3 can become a power using log rules and you will get ln y^1/3 = ln(2+x) + c

 

Now you can e both sides to cancel out ln to get

 

y^1/3 = (2+x) +e^c

 

You can get rid of the cube root by cubing both sides to get:

 

y = (2 + x + e^c)³

 

I believe this is the solution of this one. 

 

Answer 3

From e^((1/3)lny) = e^(ln(2+x)+c)...

 

e^{ln(y^(1/3))} = e^ln(2+x)e^c

y^(1/3) = e^c(2+x)

(y^(1/3))/(2+x) = e^c

ln(y^(1/3)/(2+x)) = c

 

Hope this helps!

Answer 4

Good job so far.   On the left side, notice that (1/3) ln(y) =  ln( y^1/3).     On the right side, notice that the exponential of the sum is the product of (2+x) and e^c  . But since e^c is a constant, it can be replaced with another constant - say C.

 

Then   y^1/3  =  C  (2+x)    or

 

  y =  D  (2+x)³     where D is yet another constant.   

 

The value of D would be eventually determined by an initial condition of some kind.