What Steps Can Resolve This Second-Order Differential Equation?

[PennyGirl]

Homework Statement

Solve the differential equation...
X'' = .5*x^-2
taken with respect to t
at t=0, x'=0 and x=10

Homework Equations

N/A

The Attempt at a Solution

I tried to split the variables (ie d^2 X * X^2 = .5 dt^2) but didn't get the right answer with this (i plugged it bak in and it didn't work?)
How should I start this problem?

 

[rochfor1]

Show your computations.

 

[PennyGirl]

d^2 X * X^2 = .5 dt^2
integrate both sides...
dx * x^3/3=(.5*t + C) *dt
at t=0, dx/dt=0 and x=10...
0*10^3/3 = .5*0 + C
C=0

dx * x^3/3 = .5*t*dt
integrate both sides again...
x^4 / 12 = .25*t^2 + C
same refs...
10^4/12 = .25*0 +C
C = 833.3

x^4/12 = .25*t^2 + 833.3

algebra...

x= (3*t^2 + 10000)^(1/4)

but that doesn't work...

 

[rock.freak667]

Do you know the variation of parameters method? That might work.

 

[PennyGirl]

I didn't think that would work because I thought the exponent on the x term would have to be 1, while in this case its -2...

 

[Dick]

Just integrate (1/2)x^(-2) twice. Don't forget to keep the constants of integration around.

 

[PennyGirl]

so...
integrating once...
x'=.0345*x^-2*t

then again...
x=.0345/2*x^-2*t^2+C\
?

 

[Dick]

so...
integrating once...
x'=.0345*x^-2*t

then again...
x=.0345/2*x^-2*t^2+C\
?

No, you aren't missing anything. I am. I didn't notice t was the independent variable.

 

[HallsofIvy]

d^2 X * X^2 = .5 dt^2
integrate both sides...

NO. You cannot separate a second derivative like this. A second derivative cannot be treated like a fraction.

Your original equation can be written as x^2d^2x/dt^2= 1/2[/math]. Let y= dx/dt. Then, by the chain rule, d^2x/dt= dy/dt= (dy/dx)(dx/dt)= y dy/dx. Your equation becomes y dy/dx= (1/2)x<sup>-2</sup>. Since that is now a first derivative, it <b>can</b> be treated like a fraction and separated: ydy= (1/2)x<sup>-2</sup>dx. Integrate that to find y, as a function of x, and then integrate dx/dt= y to find x.` Since you know that x(0)= 10 and x'(0)= 0, you know that y= 0 when x= 10 so can find the first constant of integration before the second integral.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> dx * x^3/3=(.5*t + C) *dt<br /> at t=0, dx/dt=0 and x=10...<br /> 0*10^3/3 = .5*0 + C<br /> C=0<br /> <br /> dx * x^3/3 = .5*t*dt<br /> integrate both sides again...<br /> x^4 / 12 = .25*t^2 + C<br /> same refs...<br /> 10^4/12 = .25*0 +C<br /> C = 833.3<br /> <br /> x^4/12 = .25*t^2 + 833.3<br /> <br /> algebra...<br /> <br /> x= (3*t^2 + 10000)^(1/4)<br /> <br /> but that doesn't work... </div> </div> </blockquote>