How Do You Solve This First Order Linear ODE?

[charmedbeauty]

Homework Statement

Solve this first order linear ODE

x (dy/dx)-2y=6x⁵

Homework Equations

The Attempt at a Solution

x (dy/dx)-2y=6x⁵

divide through by x

dy/dx -2y/x = 6x⁴

I= e^{∫-2/x dx}

=e^-ln(2) =-2

∴ -2(dy/dx) + 4y/x = -12x⁴

d/dx(-2y) = -12x⁵/5 + C

y= 6/5(x⁵) - C/2

this is wrong but where did I go wrong?

Thanks

 

[vela]

You didn't calculate the integrating factor correctly. Recheck the integral.

 

[charmedbeauty]

You didn't calculate the integrating factor correctly. Recheck the integral.

ohh it should be a half; thanks

so it should be y = y=-6/5x⁵ - 2C

the answer in my book has 2x⁵ + Ax²

 

[vela]

No, it should have an x in it.

EDIT: It refers to the integrating factor.

 

[LCKurtz]

ohh it should be a half; thanks

so it should be y = y=-6/5x⁵ - 2C

the answer in my book has 2x⁵ + Ax²

The book is correct. Show your steps and we can help you find where you went astray.

 

[Bohrok]

=e^-ln(2) =-2

This isn't true; take a look at log/exponential rules.

 

[charmedbeauty]

This isn't true; take a look at log/exponential rules.

okbthanks I figured it out now../.

I=1/x²

although I am alittle confused about this integral

if you work out the integral of -2/x

-2ln(x)

so I thought e^-2ln(x)

which is e^1/-2ln(x)


should be -1/2x

and not -1/x^2

??

thanks.

 

[HallsofIvy]

You seem to be making one algebra mistake after another. Yes, the integral of -2/x is -2ln(x) and the exponential of that is e^{-2\ln(x)} but that is NOT e/(-2\ln(x)). I can't imagine where you got that. It is, rather 1/e^{2\ln(x)}.

But what you should have done first was simplify -2 ln(x). Do you know how to take that -2 'into' the logarithm?

 

[charmedbeauty]

You seem to be making one algebra mistake after another. Yes, the integral of -2/x is -2ln(x) and the exponential of that is e^{-2\ln(x)} but that is NOT e/(-2\ln(x)). I can't imagine where you got that. It is, rather 1/e^{2\ln(x)}.

But what you should have done first was simplify -2 ln(x). Do you know how to take that -2 'into' the logarithm?

yeah I made a stupid mistake of course e^-2ln(x) = 1/e^2ln(x)

its been a while so first I should of done

-2ln(x) = -ln(x²)

Thanks