Mixing
What to multiply by to get correct form ODE
$begingroup$
Suppose $y'' + f(x)y = 0$ where $M geq f(x) geq m > 0$ on some interval $[a,b]$, then the number zeros $N$ of a non trivial solution is $lfloorfrac{(b-a)sqrt{m}}{pi}rfloor leq N leq lceilfrac{(b-a)sqrt{M}}{pi}rceil$
Simple.
Now suppose I have an equation $y''+4y'+frac{8x+sin(x)}{x+1}y = 0$ and I want to estimate the number of zeros of a non trivial solution.
I can't use the theorem as is, because the ODE is not in the correct form, to fix this, we can multiply by $e^{2x}$ and get $y''e^{2x}+4y'e^{2x}+frac{8x+sin(x)}{x+1}ye^{2x} = 0$
Now if we let $ye^{2x} = z$ we have an ODE $z'' + (frac{8x+sin(x)}{x+1}-4)z = 0$ which is in the correct form
How did the professor know to multiply by $e^{2x}$? Is there a method to this or was this just a lucky guess
calculus ordinary-differential-equations
asked Jan 29 at 17:17
Oria GruberOria Gruber
6,53732462
$endgroup$
add a comment |
$begingroup$
Suppose $y'' + f(x)y = 0$ where $M geq f(x) geq m > 0$ on some interval $[a,b]$, then the number zeros $N$ of a non trivial solution is $lfloorfrac{(b-a)sqrt{m}}{pi}rfloor leq N leq lceilfrac{(b-a)sqrt{M}}{pi}rceil$
Simple.
Now suppose I have an equation $y''+4y'+frac{8x+sin(x)}{x+1}y = 0$ and I want to estimate the number of zeros of a non trivial solution.
I can't use the theorem as is, because the ODE is not in the correct form, to fix this, we can multiply by $e^{2x}$ and get $y''e^{2x}+4y'e^{2x}+frac{8x+sin(x)}{x+1}ye^{2x} = 0$
Now if we let $ye^{2x} = z$ we have an ODE $z'' + (frac{8x+sin(x)}{x+1}-4)z = 0$ which is in the correct form
How did the professor know to multiply by $e^{2x}$? Is there a method to this or was this just a lucky guess
calculus ordinary-differential-equations
asked Jan 29 at 17:17
Oria GruberOria Gruber
6,53732462
$endgroup$
add a comment |
$begingroup$
Suppose $y'' + f(x)y = 0$ where $M geq f(x) geq m > 0$ on some interval $[a,b]$, then the number zeros $N$ of a non trivial solution is $lfloorfrac{(b-a)sqrt{m}}{pi}rfloor leq N leq lceilfrac{(b-a)sqrt{M}}{pi}rceil$
Simple.
Now suppose I have an equation $y''+4y'+frac{8x+sin(x)}{x+1}y = 0$ and I want to estimate the number of zeros of a non trivial solution.
I can't use the theorem as is, because the ODE is not in the correct form, to fix this, we can multiply by $e^{2x}$ and get $y''e^{2x}+4y'e^{2x}+frac{8x+sin(x)}{x+1}ye^{2x} = 0$
Now if we let $ye^{2x} = z$ we have an ODE $z'' + (frac{8x+sin(x)}{x+1}-4)z = 0$ which is in the correct form
How did the professor know to multiply by $e^{2x}$? Is there a method to this or was this just a lucky guess
calculus ordinary-differential-equations
asked Jan 29 at 17:17
Oria GruberOria Gruber
6,53732462
$endgroup$
Suppose $y'' + f(x)y = 0$ where $M geq f(x) geq m > 0$ on some interval $[a,b]$, then the number zeros $N$ of a non trivial solution is $lfloorfrac{(b-a)sqrt{m}}{pi}rfloor leq N leq lceilfrac{(b-a)sqrt{M}}{pi}rceil$
Simple.
Now suppose I have an equation $y''+4y'+frac{8x+sin(x)}{x+1}y = 0$ and I want to estimate the number of zeros of a non trivial solution.
I can't use the theorem as is, because the ODE is not in the correct form, to fix this, we can multiply by $e^{2x}$ and get $y''e^{2x}+4y'e^{2x}+frac{8x+sin(x)}{x+1}ye^{2x} = 0$
Now if we let $ye^{2x} = z$ we have an ODE $z'' + (frac{8x+sin(x)}{x+1}-4)z = 0$ which is in the correct form
How did the professor know to multiply by $e^{2x}$? Is there a method to this or was this just a lucky guess
calculus ordinary-differential-equations
calculus ordinary-differential-equations
asked Jan 29 at 17:17
Oria GruberOria Gruber
6,53732462
asked Jan 29 at 17:17
Oria GruberOria Gruber
6,53732462
asked Jan 29 at 17:17
Oria GruberOria Gruber
6,53732462
asked Jan 29 at 17:17
Oria GruberOria Gruber
6,53732462
asked Jan 29 at 17:17
Oria GruberOria Gruber
6,53732462
6,53732462
add a comment |
add a comment |
2 Answers
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oldest
votes
$begingroup$
If $y=ze^{g(x)}$ then we can calculate that $y'=e^{g}(z'+zg')$ and $y''=e^{g}(z''+2z'g'+zg''+z(g')^2)$.
An equation of the form $y''+A(x)y'+B(x)=0$ can then be written in terms of $z$ as $e^{g}left(z''+(2g'+A(x))z'+(g''+A(x)g'+(g')^2+B(x))right)=0$.
For this to be in the form $z''+f(x)z=0$ we need to have $2g'+A(x)=0$. So the desired substitution is $y=ze^{int A(x)/2 mathrm{d}x}$.
answered Jan 29 at 18:25
Angela RichardsonAngela Richardson
5,46011734
$endgroup$
add a comment |
$begingroup$
For me is the idea in this:
Ignore the ugly part of the equation.
What we need is to shift $y''+4y'$ so that we get $z''$. Since this first part corresponds to a linear ODE with a quadratic characteristic equation, the trick leads me to the perfect square of $r^2+4r$ which is $(r+2)^2-4.$ This says that $r=-2$ makes the job. A corresponding solution would be $e^{-2x}.$ Multiplying the initial equation by the inverse of $e^{-2x}$ makes what expected.
answered Jan 29 at 18:39
user376343user376343
3,9584829
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $y=ze^{g(x)}$ then we can calculate that $y'=e^{g}(z'+zg')$ and $y''=e^{g}(z''+2z'g'+zg''+z(g')^2)$.
An equation of the form $y''+A(x)y'+B(x)=0$ can then be written in terms of $z$ as $e^{g}left(z''+(2g'+A(x))z'+(g''+A(x)g'+(g')^2+B(x))right)=0$.
For this to be in the form $z''+f(x)z=0$ we need to have $2g'+A(x)=0$. So the desired substitution is $y=ze^{int A(x)/2 mathrm{d}x}$.
answered Jan 29 at 18:25
Angela RichardsonAngela Richardson
5,46011734
$endgroup$
add a comment |
$begingroup$
If $y=ze^{g(x)}$ then we can calculate that $y'=e^{g}(z'+zg')$ and $y''=e^{g}(z''+2z'g'+zg''+z(g')^2)$.
An equation of the form $y''+A(x)y'+B(x)=0$ can then be written in terms of $z$ as $e^{g}left(z''+(2g'+A(x))z'+(g''+A(x)g'+(g')^2+B(x))right)=0$.
For this to be in the form $z''+f(x)z=0$ we need to have $2g'+A(x)=0$. So the desired substitution is $y=ze^{int A(x)/2 mathrm{d}x}$.
answered Jan 29 at 18:25
Angela RichardsonAngela Richardson
5,46011734
$endgroup$
add a comment |
$begingroup$
If $y=ze^{g(x)}$ then we can calculate that $y'=e^{g}(z'+zg')$ and $y''=e^{g}(z''+2z'g'+zg''+z(g')^2)$.
An equation of the form $y''+A(x)y'+B(x)=0$ can then be written in terms of $z$ as $e^{g}left(z''+(2g'+A(x))z'+(g''+A(x)g'+(g')^2+B(x))right)=0$.
For this to be in the form $z''+f(x)z=0$ we need to have $2g'+A(x)=0$. So the desired substitution is $y=ze^{int A(x)/2 mathrm{d}x}$.
answered Jan 29 at 18:25
Angela RichardsonAngela Richardson
5,46011734
$endgroup$
If $y=ze^{g(x)}$ then we can calculate that $y'=e^{g}(z'+zg')$ and $y''=e^{g}(z''+2z'g'+zg''+z(g')^2)$.
An equation of the form $y''+A(x)y'+B(x)=0$ can then be written in terms of $z$ as $e^{g}left(z''+(2g'+A(x))z'+(g''+A(x)g'+(g')^2+B(x))right)=0$.
For this to be in the form $z''+f(x)z=0$ we need to have $2g'+A(x)=0$. So the desired substitution is $y=ze^{int A(x)/2 mathrm{d}x}$.
answered Jan 29 at 18:25
Angela RichardsonAngela Richardson
5,46011734
answered Jan 29 at 18:25
Angela RichardsonAngela Richardson
5,46011734
answered Jan 29 at 18:25
Angela RichardsonAngela Richardson
5,46011734
answered Jan 29 at 18:25
Angela RichardsonAngela Richardson
5,46011734
5,46011734
add a comment |
add a comment |
$begingroup$
For me is the idea in this:
Ignore the ugly part of the equation.
What we need is to shift $y''+4y'$ so that we get $z''$. Since this first part corresponds to a linear ODE with a quadratic characteristic equation, the trick leads me to the perfect square of $r^2+4r$ which is $(r+2)^2-4.$ This says that $r=-2$ makes the job. A corresponding solution would be $e^{-2x}.$ Multiplying the initial equation by the inverse of $e^{-2x}$ makes what expected.
answered Jan 29 at 18:39
user376343user376343
3,9584829
$endgroup$
add a comment |
$begingroup$
For me is the idea in this:
Ignore the ugly part of the equation.
What we need is to shift $y''+4y'$ so that we get $z''$. Since this first part corresponds to a linear ODE with a quadratic characteristic equation, the trick leads me to the perfect square of $r^2+4r$ which is $(r+2)^2-4.$ This says that $r=-2$ makes the job. A corresponding solution would be $e^{-2x}.$ Multiplying the initial equation by the inverse of $e^{-2x}$ makes what expected.
answered Jan 29 at 18:39
user376343user376343
3,9584829
$endgroup$
add a comment |
$begingroup$
For me is the idea in this:
Ignore the ugly part of the equation.
What we need is to shift $y''+4y'$ so that we get $z''$. Since this first part corresponds to a linear ODE with a quadratic characteristic equation, the trick leads me to the perfect square of $r^2+4r$ which is $(r+2)^2-4.$ This says that $r=-2$ makes the job. A corresponding solution would be $e^{-2x}.$ Multiplying the initial equation by the inverse of $e^{-2x}$ makes what expected.
answered Jan 29 at 18:39
user376343user376343
3,9584829
$endgroup$
For me is the idea in this:
Ignore the ugly part of the equation.
What we need is to shift $y''+4y'$ so that we get $z''$. Since this first part corresponds to a linear ODE with a quadratic characteristic equation, the trick leads me to the perfect square of $r^2+4r$ which is $(r+2)^2-4.$ This says that $r=-2$ makes the job. A corresponding solution would be $e^{-2x}.$ Multiplying the initial equation by the inverse of $e^{-2x}$ makes what expected.
answered Jan 29 at 18:39
user376343user376343
3,9584829
answered Jan 29 at 18:39
user376343user376343
3,9584829
answered Jan 29 at 18:39
user376343user376343
3,9584829
answered Jan 29 at 18:39
user376343user376343
3,9584829
3,9584829
add a comment |
add a comment |
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