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[SOLVED] second order homogeneous equations with non constant coefficients

shorty

New member
Feb 5, 2012
16
I was given a question and i am really unsure how to go about solving it. it appears to be solveable using the characteristic equation and whatnot, however i have my coeffecients in terms of the independent variable. so i am confused. the question initially asked to compute the wronskian, and it gave the solutions, then it gives initial values to solve. There's where the confusion began. I'd be grateful for some guidance please.

Solve the initial value problem
\(\displaystyle 2t^2y''+3ty'-y= 0; y(1)=2, y'(1)=1\), given that \(\displaystyle y_1(t)=\sqrt(t) \mbox{ and } y_2​(t)=\frac{1}{t}\)


The question also asked to use the Wronskian to show that the two solutions are linearly independent, then said, hence, solve the ivp (given above). I calculated the Wronskian to be \(\displaystyle \frac{-3 \sqrt{t}}{2t^2​}\)



(These delimiters are not working for me, they are doing their own thing. quite annoying. )
 
Last edited by a moderator:

dwsmith

Well-known member
Feb 1, 2012
1,673
I was given a question and i am really unsure how to go about solving it. it appears to be solveable using the characteristic equation and whatnot, however i have my coeffecients in terms of the independent variable. so i am confused. the question initially asked to compute the wronskian, and it gave the solutions, then it gives initial values to solve. There's where the confusion began. I'd be grateful for some guidance please.

Solve the initial value problem
$$ 2t2y''+3ty'-y= 0; y(1)=2, y'(1)=1, given that y1(t)=\sqrt(t) and y2​(t)=\frac{1}{t}


The question also asked to use the Wronskian to show that the two solutions are linearly independent, then said, hence, solve the ivp (given above). I calculated the Wronskian to be \frac{-3 \sqrt{t}}{2t2​} $$



(These delimiters are not working for me, they are doing their own thing. quite annoying. )
You need to start what you want in math with $$ and then close it in with it \$\$ no slashes of course at the end.
 

shorty

New member
Feb 5, 2012
16
isn't that what i've done?

You need to start what you want in math with $$ and then close it in with it \$\$ no slashes of course at the end.
 

dwsmith

Well-known member
Feb 1, 2012
1,673

shorty

New member
Feb 5, 2012
16
well your 'iput' came out as Math. same thing was happening to part of what i was writing, and not the actual math that i wanted in the latex.

$$ 2t2y''+3ty'-y= 0; y(1)=2, y'(1)=1, \ \text{given that} \ y1(t)=\sqrt(t) and y2​(t)=\frac{1}{t} $$

I put $$ at the end of this math.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
well your 'iput' came out as Math. same thing was happening to part of what i was writing, and not the actual math that i wanted in the latex.
$$
2t^2y''+3ty'-y= 0; y(1)=2, y'(1)=1, \ \text{given that} \ y_1(t)=\sqrt{t} and y_2(t)=\frac{1}{t}
$$

Will this work?
 

shorty

New member
Feb 5, 2012
16
neither. i tried that too. is it me? i just restarted my browser and getting the same thing. the thing is when i started writing the question i previewed just as i started and the math came out fine, then when i added more it messed up and that was it.

$$
2t^2y''+3ty'-y= 0; y(1)=2, y'(1)=1, given that y_1(t)=\sqrt(t) and y_2(t)=\frac{1}{t}
$$

Will this work?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
neither. i tried that too. is it me? i just restarted my browser and getting the same thing. the thing is when i started writing the question i previewed just as i started and the math came out fine, then when i added more it messed up and that was it.
Hi shorty,

I think you want to solve,

\[2t^{2}y''+3ty'-y= 0~;~y(1)=2,~ y'(1)=1,~\mbox{given that }y_{1}(t)=\sqrt{t}\mbox{ and }y_{2}​(t)=\frac{1}{t}\]

These type of differential equations are known as Cauchy-Euler equations. Although your problem is to find the Wronskian, solving these kind of equations are quite easy. I will solve it just for the sake of personal gratification. (Happy)

Substitute, $y=t^m$ and you get,

\[2m(m-1)t^{m}+3mt^{m}-t^{m}=0\]

\[\Rightarrow 2m(m-1)+3m-1=0\]

\[\Rightarrow 2m^2+m-1=0\]

\[\Rightarrow m=0.5,-1\]

Therefore, $y_{1}=t^{\frac{1}{2}}$ and $y_{2}=t^{-1}$. And the given solutions are correct. Now we shall determine the Wronskian,

\[W(t^{\frac{1}{2}},t^{-1})=\begin{vmatrix}t^{\frac{1}{2}}& t^{-1}\\ \frac{1}{2}t^{-\frac{1}{2}}& -t^{-2}\end{vmatrix}=-t^{-\frac{3}{2}}-\frac{1}{2}t^{-\frac{3}{2}}=-\frac{3}{2}t^{-\frac{3}{2}}\]

So $W(t^{\frac{1}{2}},t^{-1})\neq 0$. Hence the two solutions are linearly independent. Write the general solution of the system and find the two constants using the given initial values. I hope you can continue. (Happy)
 

shorty

New member
Feb 5, 2012
16
thanks,

but this is not what i need, i got this part already myself. my problem is the part after.. i don't kno how to calculate this with the info given.. all the examples i have seen don't show how to do it.. the coefficients are not constant, so the auxiliary equation doesn't work.. i don't kno how to incorporate the wronkskian and the solutions to solve the ivp..

also how did your math show up? my attempts proved futile. $'s, \ and ['s , and even \ and ('s ... nothing...
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
I was given a question and i am really unsure how to go about solving it. it appears to be solveable using the characteristic equation and whatnot, however i have my coeffecients in terms of the independent variable. so i am confused. the question initially asked to compute the wronskian, and it gave the solutions, then it gives initial values to solve. There's where the confusion began. I'd be grateful for some guidance please.

Enclose only the equations with the $$, $, \( or \[ symbols. If you want to insert text within equations use \mbox{} command. I have cleaned your LaTeX code below.

Solve the initial value problem
$$2t^{2}y''+3ty'-y= 0;~ y(1)=2,~ y'(1)=1,\mbox{ given that }y_{1}(t)=\sqrt{t}\mbox{ and }y_{2}​(t)=\frac{1}{t}$$


The question also asked to use the Wronskian to show that the two solutions are linearly independent, then said, hence, solve the ivp (given above). I calculated the Wronskian to be $\dfrac{-3}{2\sqrt{t^3}}$.



(These delimiters are not working for me, they are doing their own thing. quite annoying.)

Now they are working aren't they? (Happy)
thanks,

but this is not what i need, i got this part already myself. my problem is the part after.. i don't kno how to calculate this with the info given.. all the examples i have seen don't show how to do it.. the coefficients are not constant, so the auxiliary equation doesn't work.. i don't kno how to incorporate the wronkskian and the solutions to solve the ivp..

also how did your math show up? my attempts proved futile. $'s, \ and ['s , and even \ and ('s ... nothing...
With reference to my previous post you can write the general solution as,

\[y(t)=C_{1}\sqrt{t}+C_{2}t^{-1}\mbox{ where }C_{1}\mbox{ and }C_{2}\mbox{ are constants.}\]

So you are given that, $y(1)=2\mbox{ and }y'(1)=1$. Substituting these values in the general solution you can get two equations from which you can find $C_{1}$ and $C_{2}$. Now I am sure you can proceed.
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,041
(These delimiters are not working for me, they are doing their own thing. quite annoying. )
Hi there. You're using some strange syntax for exponents and subscripts. Use ^ for exponents and _ for subscript.
 

shorty

New member
Feb 5, 2012
16
Oh, thanks,

i was using the font options enclosed in the text box. just where you choose the font and other options above where you type. there are sub and superscript tabs.

thanks, i will not use those in future.
Hi there. You're using some strange syntax for exponents and subscripts. Use ^ for exponents and _ for subscript.


---------- Post added at 06:29 ---------- Previous post was at 06:29 ----------

Ohhhh... how straightforward now.

thank you so very much.


With reference to my previous post you can write the general solution as,

\[y(t)=C_{1}\sqrt{t}+C_{2}t^{-1}\mbox{ where }C_{1}\mbox{ and }C_{2}\mbox{ are constants.}\]

So you are given that, $y(1)=2\mbox{ and }y'(1)=1$. Substituting these values in the general solution you can get two equations from which you can find $C_{1}$ and $C_{2}$. Now I am sure you can proceed.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
Note also that you can solve any Cauchy-Euler equation by performing the substitution $x=\ln(t)$. This transforms your equation into a second-order linear ODE with constant coefficients.

Note also that $\LaTeX$ does not appear to be showing up correctly in Internet Explorer. We are working on this.