Differential Equations reduction of order

[iamtrojan3]

Homework Statement

Find y2 given
t^2y'' + 3ty' + y = 0 y1(t) = 1/t

Homework Equations

The Attempt at a Solution

i found y, y' and y'' and plugged it back into the Differential equation,
after doing some work, I'm basicly down to
tv''+v' = 0
where y2 = v/t

so u = v' and u' = v'' Plugging that in...
tu' + u = 0
how would i solve this for u?
i think its separable equations but that would give me
du/-u = dt/t
which would be ln (u) = -ln(t) + c ... u = 1/t +c so v' = 1/t +c which means v = lnt
pluging v into y2 would give me lnt/t
Sorry its kind of confusing but the answer's not in the back of the book and i need to know if i did this right

Thank you!

 

[HallsofIvy]

Homework Statement

Find y2 given
t^2y'' + 3ty' + y = 0 y1(t) = 1/t

Homework Equations

The Attempt at a Solution

i found y, y' and y'' and plugged it back into the Differential equation,
after doing some work, I'm basicly down to
tv''+v' = 0
where y2 = v/t

so u = v' and u' = v'' Plugging that in...
tu' + u = 0
how would i solve this for u?
i think its separable equations but that would give me
du/-u = dt/t
which would be ln (u) = -ln(t) + c ... u = 1/t +c so v' = 1/t +c which means v = lnt
pluging v into y2 would give me lnt/t
Sorry its kind of confusing but the answer's not in the back of the book and i need to know if i did this right

Thank you!

Yes, the problem reduces to tu'+ u= 0 which is separable: du/u= -dt/t. Integrating ln(u)= -ln(t)+ c and, taking the exponential of both sides u= C/t where C= e^c. (Notice that the C is multiplied, not added: e<sup>a+c[/b]= eaec, not "ea+ c".)

The integral of u= v'= C/t is v(t)= Cln(t)+ D and then, since you made the reduction by assuming a solution of the form y= v(t)y1= v(t)(1/t), your general solution to the original differential equation is y(t)= Cln(t)/t+ D/t. How does that compare with the solution in the back of your book?</sup>