Hello!

I have problem:

Derive the solution of the ordinary differential equation

d2y/dx2 =f(x),x>0, y(x)=0, dy/dx (0)=0,

in form y(x)= integral from 0 to x [(x-t)f(t)dt].

tnxs

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- Aug 13th 2007, 07:33 AMperfect[SOLVED] I need help for derive d2y/dx2 in integral form
Hello!

I have problem:

Derive the solution of the ordinary differential equation

**d2y/dx2 =f(x),**x>0, y(x)=0, dy/dx (0)=0,

in form y(x)= integral from 0 to x [(x-t)f(t)dt].

tnxs - Aug 13th 2007, 09:48 AMCaptainBlack
Some clarification is needed. It is not possible in general to write the solution of:

$\displaystyle

\frac{d^2y}{dx^2}=f(x)

$

with $\displaystyle y(0)=0,\ y'(0)=0$, in the form

$\displaystyle y(x)=\int_0^x (x-t) f(t) dt$.

To show this just put $\displaystyle f(x)=x$.

RonL - Aug 16th 2007, 11:48 PMRebesques
perfect:

Quote:

Derive the solution of the ordinary differential equation

d2y/dx2 =f(x), x>0, y(x)=0, dy/dx (0)=0,

in form y(x)= integral from 0 to x [(x-t)f(t)dt]

Are you asked to actually prove the formula, or just show this is the solution? Because the latter is easy: Differentiate twice (under the integral sign) to get y"=f, and since y also gives y(0)=0, y'(0)=0, from the uniqueness theorem this is the only solution!

Yes, I know it's cheating, but it's a way. Now if you are asked to derive the formula, write

$\displaystyle y(x)=y(x)-y(0)=\int_{0}^{x}y'(t)dt$

and integrating by parts,

$\displaystyle y(x)=\int_{0}^{x}y'(t)(t)'dt=y'(t)t\bigg|_{0}^x-\int_0^xy''(t)tdt=xy'(x)-\int_0^xf(t)tdt$, (1)

since y''=f. Now one integration of the differential equation gives us $\displaystyle y'(x)=\int_0^xf(t)dt$, and substitute into (1) to get

$\displaystyle y(x)=x\int_0^xf(t)dt-\int_0^xf(t)tdt=\int_0^x(x-t)f(t)dt$.

Captainblack:

Quote:

To show this just put...

- Aug 17th 2007, 10:49 AMCaptainBlackQuote:

Captainblack:

Quote:

To show this just put...

RonL