# Thread: [SOLVED] I need help for derive d2y/dx2 in integral form

1. ## [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

2. Originally Posted by perfect
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
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

3. perfect:
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:
To show this just put...
The function must also satisfy the initial conditions.

4. Captainblack:
To show this just put...
The function must also satisfy the initial conditions.
In fact I was misreading the question statement in another way (the function does not have to satisfy the initial condition but the solution does).

RonL

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