Solve Math: Step-by-Step Guide to Finding y = cx^2

[cscott]

Can anyone walk me through solving this?

$$\frac{dy}{dx} = \frac{2y}{x}$$
separate veriables...
$$\frac{dy}{2y} = \frac{dx}{x}$$
...and integrate from here, but the answer is $y = cx^2$ and I don't see how they got that.

 

[TD]

Show us how you integrated, it shouldn't be so hard

 

[cscott]

Show us how you integrated, it shouldn't be so hard

Ya, I know

Can't I say that it's equal to
$$\int \frac{1}{2}y^{-1}dx =\int \frac{1}{x} dx$$
...isn't the integral of $\frac{1}{x} = ln (|x|)$?

 

[d_leet]

Ya, I know

Can't I say that it's equal to
$$\int \frac{1}{2}y^{-1}dx =\int \frac{1}{x} dx$$
...isn't the integral of $\frac{1}{x} = ln (|x|)$?

Yes, that looks right so far.

 

[mathmike]

the left side of your equation should be inegrated with respect to y not x

 

[cscott]

the left side of your equation should be inegrated with respect to y not x

Sorry, that was a typo.

So I should get 1/2 ln(|y|) + C' = ln(|x|) + C after I integrate? Hows that equal to y =cx^2? :\

 

[nocturnal]

combine C' and C into one constant, D, and refer to the properties of natural log

 

[cscott]

combine C' and C into one constant, D, and refer to the properties of natural log

C - C' = D

$$\ln y = \ln x^2 + D$$

and I can just remove ln, even with D there?

 

[nocturnal]

and I can just remove ln, even with D there?

no, use the properties of ln to simplify your equation.

hint:
$$\ln a - \ln b = \ \ ??$$

$$e^{\ln a} = \ \ ??$$

$$b \ln a = \ \ ??$$

 

[cscott]

So,

$$\ln \frac{y}{x^2} = D$$
$$e^{\ln y/x^2} = e^D$$

But then wouldn't I end up with

$$y = e^Dx^2$$

 

[TD]

Yes, but since e is a constant and D is too, e^d is just another constant! Name it c and you have what you want

 

[cscott]

Yes, but since e is a constant and D is too, e^d is just another constant! Name it c and you have what you want

Oh, well that makes sense. Thanks everyone.