Simple Differential equation with reduction of order

[trap101]

use method of reduction of order to find second solution:

t²y''-4ty+6y = 0 , y₁(t)= t²


Attempt:

So I've done all the steps, up to the substitution, but I'm having problems with what appears to be a simple linear equation but I can't solve it: Any ways, with w = v' I arrive at:

w't⁴ = 0

Everything else canceled out. But trying to get an integrating factor with this would lead me to 0.

 

[LCKurtz]

use method of reduction of order to find second solution:

t²y''-4ty+6y = 0 , y₁(t)= t²Attempt:

So I've done all the steps, up to the substitution, but I'm having problems with what appears to be a simple linear equation but I can't solve it: Any ways, with w = v' I arrive at:

w't⁴ = 0

Everything else canceled out. But trying to get an integrating factor with this would lead me to 0.

Actually, you are OK so far. Divide out the $t^4$ and you have $w'=0$ or $v''=0$. Integrate it twice and remember when you are done that $y=t^2v$. You will see what a second independent solution is.

 

[trap101]

Actually, you are OK so far. Divide out the $t^4$ and you have $w'=0$ or $v''=0$. Integrate it twice and remember when you are done that $y=t^2v$. You will see what a second independent solution is.

Your right. This working with zero in this scenario is so weird...

I had another quick question with regards to undetermined coefficients:

y''+2y' = 3 + 4sin(2t)

So working out everything I get everything correct, but I'm missing a term in the particular solution. my solution is:

Y_p(t) = -1/2cos(2t) - 1/2sin(2t)

I guess the issue is when I'm setting up my guess and setting the expressions equal. so my initial "guess" is:

Y_p(t) = A + B cos(2t) + C sin(2t)

obviously after diffeentiating this twice the constant A term disappeared and as well there is no "y" in the differential equation. yet in the solution they got:

3/2 -1/2cos(2t) - 1/2sin(2t) plus the complimentary equation. How do I get the 3/2?

 

[LCKurtz]

Your right. This working with zero in this scenario is so weird...

I had another quick question with regards to undetermined coefficients:

y''+2y' = 3 + 4sin(2t)

So working out everything I get everything correct, but I'm missing a term in the particular solution. my solution is:

Y_p(t) = -1/2cos(2t) - 1/2sin(2t)

I guess the issue is when I'm setting up my guess and setting the expressions equal. so my initial "guess" is:

Y_p(t) = A + B cos(2t) + C sin(2t)

obviously after diffeentiating this twice the constant A term disappeared and as well there is no "y" in the differential equation. yet in the solution they got:

3/2 -1/2cos(2t) - 1/2sin(2t) plus the complimentary equation. How do I get the 3/2?

You didn't show your $y_c$ complementary solution. But $y=A$ obviously is a solution of the homogeneous equation, so it can't work for $y_p$. Try $y_p=At$ plus the sine an cosine terms.

 

[trap101]

You didn't show your $y_c$ complementary solution. But $y=A$ obviously is a solution of the homogeneous equation, so it can't work for $y_p$. Try $y_p=At$ plus the sine an cosine terms.

Sorry for the delayed response. had class. my complementary solution was:

y_c(t) = c₁e^0t + c₂e^-2t

So where does the A term appear in the complemetary solution? is it because c₁ is only the constant without e^0t in it?

 

[LCKurtz]

Sorry for the delayed response. had class. my complementary solution was:

y_c(t) = c₁e^0t + c₂e^-2t

So where does the A term appear in the complemetary solution? is it because c₁ is only the constant without e^0t in it?

If you simplify $c_1e^{0t}$ what do you get?

 

[trap101]

If you simplify $c_1e^{0t}$ what do you get?

Yes. I get the constant alone. Is it because it's the constant or should I look at it as it's because they are the same degree polynomial?

 

[LCKurtz]

Yes. I get the constant alone. Is it because it's the constant or should I look at it as it's because they are the same degree polynomial?

Too many pronoun "its" in that question. I have no idea what you are asking me.

 

[trap101]

Too many pronoun "its" in that question. I have no idea what you are asking me.

Sorry. What I was asking is that since in the complementary solution I have y(t) = c₁ + c₂e^-2 does the constant c₁ match with the "A" that I have as my "guess" for the polynomial portion of the particular solution? and if so, then that's my I add a "t" to it? i.e: At + (the rest)?

 

[LCKurtz]

Sorry. What I was asking is that since in the complementary solution I have y(t) = c₁ + c₂e^-2 does the constant c₁ match with the "A" that I have as my "guess" for the polynomial portion of the particular solution? and if so, then that's my I add a "t" to it? i.e: At + (the rest)?

Yes. The fact that a constant $c_1$ is a solution of the homogeneous equation means that a constant, no matter whether you call it $A$ or something else can't be a solution of the NH equation. So no point in trying $y_p = A$.