How do I show that a particular solution $y_{1}$ of the ODE $$y''-k^{2}y=R(x)$$ $k\neq0$, is given by $$y_{1}=\frac{1}{k}\int_{0}^{x}R(t)\sinh(k(x-t))dt$$
I really have no idea how to do it.
Asked
Active
Viewed 167 times
2
-
Three things: (i) you must find out what it means for $y_1$ to be a solution of $y'' - k^2y = R(x)$, (ii) you must know the fundamental theorem of calculus and (iii) $\sinh(x) = \frac{1}{2}(e^x - e^{-x}).$ – snar Jul 09 '14 at 05:55
-
@snarski you can't readily apply the FTC here, the integrand depends on $x$ as well. – DanZimm Jul 09 '14 at 06:01
-
2see http://math.stackexchange.com/questions/860604/prove-that-a-function-is-solution-of-ivp – Leucippus Jul 09 '14 at 06:02
-
What are the intial conditions? – Mhenni Benghorbal Jul 09 '14 at 06:09
2 Answers
2
The full version of the general solution of the corresponding ODE done by variation of parameters is $y=C_1\sinh kx+C_2\cosh kx+\sinh kx\int_0^x\dfrac{R(x)\cosh kx}{k}dx-\cosh kx\int_0^x\dfrac{R(x)\sinh kx}{k}dx$
Note that the general solution can also rewrite as
$y=C_1\sinh kx+C_2\cosh kx+\dfrac{\sinh kx}{k}\int_0^xR(t)\cosh kt~dt-\dfrac{\cosh kx}{k}\int_0^xR(t)\sinh kt~dt$
$y=C_1\sinh kx+C_2\cosh kx+\dfrac{1}{k}\int_0^xR(t)(\sinh kx\cosh kt-\cosh kx\sinh kt)~dt$
$y=C_1\sinh kx+C_2\cosh kx+\dfrac{1}{k}\int_0^xR(t)\sinh(k(x-t))~dt$
The particular solution part is $y_1=\dfrac{1}{k}\int_0^xR(t)\sinh(k(x-t))~dt$
doraemonpaul
- 16,488
0
Hint: You can use the Laplace transform techniques to get this solution.
Added: You can follow the steps:
1) Take the Laplace transform of the ode.
2) Rearrange and solve for $Y(s)$
3) apply the initial conditions if they have been given
4) take the inverse Laplace transform and you need to use the convolution property of the Laplace transform.
Mhenni Benghorbal
- 48,099