very differential equation with not known source function

logistic_guy said:
thank

how to do that?
Click to expand...
In response #38 you have calculated the general form of u(x) and you want to calculate the validity of that solution. Given bdy conditions are

u(0)=u(1)=0

So calculate u(0) and u(1) → do those match given bdy. condition u(0)=u(1)=0? .....................................(1)

If those do not match - then you have made a mistake somewhere. In that case, start over.

Now derive u'(x) and u"(x) ← use those in given original DE . Does u(x) satisfy the given DE? .............(2)

If the answers to both (1) & (2) are affirmative, then most probably your solution is correct.
 
logistic_guy said:
homegenous solution
\(\displaystyle u(x) = c_1e^{2x} + c_2e^{-2x}\)
particular solution
\(\displaystyle u(x) = a_0\)
\(\displaystyle 0 - 4u_0 = 1\)
\(\displaystyle u_0 = \frac{-1}{4}\)
general solution
\(\displaystyle u(x) = c_1e^{2x} + c_2e^{-2x} - \frac{1}{4}\)
first condition
\(\displaystyle 0 = c_1e^{0} + c_2e^{0} - \frac{1}{4}\)
\(\displaystyle c_1 + c_2 = \frac{1}{4}\)
second condition
\(\displaystyle 0 = c_1e^{2} + c_2e^{-2} - \frac{1}{4}\)
\(\displaystyle c_2 = \frac{1}{4} - c_1\)
\(\displaystyle 0 = c_1e^{2} + (\frac{1}{4} - c_1)e^{-2} - \frac{1}{4}\)
\(\displaystyle 0 = c_1e^{2} + \frac{1}{4}e^{-2} - c_1e^{-2} - \frac{1}{4}\)
\(\displaystyle -\frac{1}{4}e^{-2} + \frac{1}{4} = c_1e^{2} - c_1e^{-2}\)
\(\displaystyle \frac{-\frac{1}{4}e^{-2} + \frac{1}{4}}{e^{2} - e^{-2}} = c_1\)
\(\displaystyle c_2 = \frac{1}{4} - \frac{-\frac{1}{4}e^{-2} + \frac{1}{4}}{e^{2} - e^{-2}}\)
\(\displaystyle u(x) = (\frac{-\frac{1}{4}e^{-2} + \frac{1}{4}}{e^{2} - e^{-2}})e^{2x} + (\frac{1}{4} - \frac{-\frac{1}{4}e^{-2} + \frac{1}{4}}{e^{2} - e^{-2}})e^{-2x} - \frac{1}{4}\)
Click to expand...
According to Mrs. Alpha, the solution is:

[imath]\displaystyle u(x) = -\frac{-e^{2-2x}-e^{2x} + 1 + e^{2}}{4 + 4e^{2}}[/imath]

Let us compare this with your solution. According to Mrs. Alpha, your solution is:

[imath]\displaystyle u(x) = \left(\frac{-\frac{1}{4}e^{-2} + \frac{1}{4}}{e^{2} - e^{-2}}\right)e^{2x} + \left(\frac{1}{4} - \frac{-\frac{1}{4}e^{-2} + \frac{1}{4}}{e^{2} - e^{-2}}\right)e^{-2x} - \frac{1}{4} = \frac{e^{2x} + e^{2(1-x)} - 1 - e^{2}}{4(1 + e^{2})}[/imath]

which is the same as the first solution. Bravo👏

Now, let us use our green function solution in post #31. Notice that [imath]a = 0, b = 1, \ \text{and} \ f(s) = 1[/imath]. According to Mrs. Alpha, the solution is:

[imath]\displaystyle u(x) = \int_{a}^{b} G(x,s) f(s) \ ds = \int_{0}^{1} G(x,s) \ ds[/imath]

[imath]\displaystyle = -\frac{1}{2\sinh(2[1 - 0])} \left(\int_{0}^{x} \sinh(2[1 - x])\sinh(2[s - 0]) \ ds + \int_{x}^{1} \sinh(2[x - 0])\sinh(2[1 - s]) \ ds\right)[/imath]


[imath]\displaystyle = \frac{(e^{x} - e^{2 - x}) \sinh(x)}{2 (1 + e^{2})} = \frac{e^{2-2x} + e^{2x} - 1 - e^{2}}{4(1 + e^{2})}[/imath]

You can even solve this integral by hand if you want. The conclusion is that our green function solution is correct. If you notice that our green function solution [imath]G(x,s)[/imath] is independent from the source function [imath]f(s)[/imath] which makes it so powerful to solve instantly any nonhomogeneous differential equation of this type [imath]u''(x) - k^2u(x) = f(x)[/imath], [imath]a < x < b[/imath].

It will take some time to understand what this does mean, but this is the main idea of Green Function in differential equations.
 
khansaheb said:
Using ALGEBRA - most of the time!!
Click to expand...
i'm very good in algebra:)

khansaheb said:
In response #38 you have calculated the general form of u(x) and you want to calculate the validity of that solution. Given bdy conditions are

u(0)=u(1)=0

So calculate u(0) and u(1) → do those match given bdy. condition u(0)=u(1)=0? .....................................(1)

If those do not match - then you have made a mistake somewhere. In that case, start over.

Now derive u'(x) and u"(x) ← use those in given original DE . Does u(x) satisfy the given DE? .............(2)

If the answers to both (1) & (2) are affirmative, then most probably your solution is correct.
Click to expand...
can you please show me how you'll do it to post 31?

mario99 said:
According to Mrs. Alpha, the solution is:

[imath]\displaystyle u(x) = -\frac{-e^{2-2x}-e^{2x} + 1 + e^{2}}{4 + 4e^{2}}[/imath]

Let us compare this with your solution. According to Mrs. Alpha, your solution is:

[imath]\displaystyle u(x) = \left(\frac{-\frac{1}{4}e^{-2} + \frac{1}{4}}{e^{2} - e^{-2}}\right)e^{2x} + \left(\frac{1}{4} - \frac{-\frac{1}{4}e^{-2} + \frac{1}{4}}{e^{2} - e^{-2}}\right)e^{-2x} - \frac{1}{4} = \frac{e^{2x} + e^{2(1-x)} - 1 - e^{2}}{4(1 + e^{2})}[/imath]

which is the same as the first solution. Bravo👏

Now, let us use our green function solution in post #31. Notice that [imath]a = 0, b = 1, \ \text{and} \ f(s) = 1[/imath]. According to Mrs. Alpha, the solution is:

[imath]\displaystyle u(x) = \int_{a}^{b} G(x,s) f(s) \ ds = \int_{0}^{1} G(x,s) \ ds[/imath]

[imath]\displaystyle = -\frac{1}{2\sinh(2[1 - 0])} \left(\int_{0}^{x} \sinh(2[1 - x])\sinh(2[s - 0]) \ ds + \int_{x}^{1} \sinh(2[x - 0])\sinh(2[1 - s]) \ ds\right)[/imath]


[imath]\displaystyle = \frac{(e^{x} - e^{2 - x}) \sinh(x)}{2 (1 + e^{2})} = \frac{e^{2-2x} + e^{2x} - 1 - e^{2}}{4(1 + e^{2})}[/imath]

You can even solve this integral by hand if you want. The conclusion is that our green function solution is correct. If you notice that our green function solution [imath]G(x,s)[/imath] is independent from the source function [imath]f(s)[/imath] which makes it so powerful to solve instantly any nonhomogeneous differential equation of this type [imath]u''(x) - k^2u(x) = f(x)[/imath], [imath]a < x < b[/imath].

It will take some time to understand what this does mean, but this is the main idea of Green Function in differential equations.
Click to expand...
thank very much mario99

i think i'm understanding. your solution is correct. can you please show me variation of parameters?🫣
 
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