Hi! I am trying to solve the following question and this is what I have so far, If someone could help me with the following steps I would greatly appreciate it!!! =D
Q: Solve the equation t^2*y" -2*y = 3*t^2-1 by using the fact that y1=t^-1 is one solution of the corresponding homogeneous equation.
First apply reduction of order for finding the second solution of the homogeneous equation:
let y = v * t^-1
take the first derivative: y'= v' *t^-1 -t^-2*v
take the second derivative: y"= (v" * t^-1) - (2*t^-2*v')+(2*t^-2*v)
Substitute into original equation and simplify to obtain: v"-2*v'+2*v-2*t =0
I am stuck on this part... help?
By undetermined coefficients the first partial solution has the form: At+B
The second partial solution simply has -1 as a solution (in order to add later by the theorem of fundamental set of solutions and what-not...
Is this correct so far? How do I complete it? If anyone can help step by step I would really appreciate it! =)) Thanks!
Q: Solve the equation t^2*y" -2*y = 3*t^2-1 by using the fact that y1=t^-1 is one solution of the corresponding homogeneous equation.
First apply reduction of order for finding the second solution of the homogeneous equation:
let y = v * t^-1
take the first derivative: y'= v' *t^-1 -t^-2*v
take the second derivative: y"= (v" * t^-1) - (2*t^-2*v')+(2*t^-2*v)
Substitute into original equation and simplify to obtain: v"-2*v'+2*v-2*t =0
I am stuck on this part... help?
By undetermined coefficients the first partial solution has the form: At+B
The second partial solution simply has -1 as a solution (in order to add later by the theorem of fundamental set of solutions and what-not...
Is this correct so far? How do I complete it? If anyone can help step by step I would really appreciate it! =)) Thanks!
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Let y = t^(-1) * v.
So, y' = -t^(-2) v + t^(-1) v'
y'' = 2t^(-3) v - 2t^(-2) v' + t^(-1) v''.
Substituting this into the DE yields
t^2 [2t^(-3) v - 2t^(-2) v' + t^(-1) v''] - 2(t^(-1) v) = 3t^2 - 1
==> tv'' - 2v' = 3t^2 - 1; note that this is first order linear in v'.
Next, rewrite in standard form (leading coefficient 1):
v'' + (-2/t)v' = 3t - 1/t.
Multiply both sides by the integrating factor
e^[∫ (-2/t) dt] = e^(-2 ln t) = e^(ln(t^(-2)) = t^(-2):
t^(-2) v'' - 2t^(-3) v' = 3t^(-1) - t^(-3)
==> (d/dt) [t^(-2) v'] = 3t^(-1) - t^(-3).
Integrate both sides:
t^(-2) v' = 3 ln t + (1/2) t^(-2) + A
==> v' = 3t^2 ln t + 1/2 + At^2.
Integrate both sides again (and use a little integration by parts):
v = (t^3 ln t - (1/3)t^3) + (1/2)t + (A/3)t^3 + B
==> y = t^(-1) v = t^2 ln t + 1/2 + (A/3 - 1/3)t^2 + Bt^(-1).
Relabeling the constants, we can write the general solution as
y = C₁ t^2 + C₂ t^(-1) + (t^2 ln t + 1/2).
Double check:
http://www.wolframalpha.com/input/?i=t^2…
I hope this helps!
So, y' = -t^(-2) v + t^(-1) v'
y'' = 2t^(-3) v - 2t^(-2) v' + t^(-1) v''.
Substituting this into the DE yields
t^2 [2t^(-3) v - 2t^(-2) v' + t^(-1) v''] - 2(t^(-1) v) = 3t^2 - 1
==> tv'' - 2v' = 3t^2 - 1; note that this is first order linear in v'.
Next, rewrite in standard form (leading coefficient 1):
v'' + (-2/t)v' = 3t - 1/t.
Multiply both sides by the integrating factor
e^[∫ (-2/t) dt] = e^(-2 ln t) = e^(ln(t^(-2)) = t^(-2):
t^(-2) v'' - 2t^(-3) v' = 3t^(-1) - t^(-3)
==> (d/dt) [t^(-2) v'] = 3t^(-1) - t^(-3).
Integrate both sides:
t^(-2) v' = 3 ln t + (1/2) t^(-2) + A
==> v' = 3t^2 ln t + 1/2 + At^2.
Integrate both sides again (and use a little integration by parts):
v = (t^3 ln t - (1/3)t^3) + (1/2)t + (A/3)t^3 + B
==> y = t^(-1) v = t^2 ln t + 1/2 + (A/3 - 1/3)t^2 + Bt^(-1).
Relabeling the constants, we can write the general solution as
y = C₁ t^2 + C₂ t^(-1) + (t^2 ln t + 1/2).
Double check:
http://www.wolframalpha.com/input/?i=t^2…
I hope this helps!
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beats me