Question

An integral equation is an equation that contains an unknown function y(x) and an integral that involves y(x). Solve the given integral equation. [Hint: Use an initial condition obtained from the integral equation.] y(x) = 2 + x [t − ty(t)] dt 2

Answer 1

Answer:

$y(x)=e^{\frac{-x^{2}}{2}+2 }+1$

Step-by-step explanation:

We have that:

$y(x)=2+\int\limits^x_2 {[t-ty(t)]} \, dt$      (Equation 1)

To resolve this integral equation, we need to use the second Fundamental Theorem of Calculus, which says:

$\frac{d}{dx} [\int\limits^x_a {f(t)} \, dt]=f(x)$

So, we need to differentiate both sides of equation 1 with respect to x:

$\frac{dy}{dx} =\frac{d}{dx} [2+\int\limits^x_2 {[t-ty(t)]} \, dt]$

$\frac{dy}{dx} =\frac{d(2)}{dx}+\frac{d}{dx} [\int\limits^x_2 {[t-ty(t)]} \, dt]$

We know that the derivate for a constant value is zero. And,

$\frac{dy}{dx}=\frac{d}{dx} [\int\limits^x_2 {t} \, dt]-\frac{d}{dx}[\int\limits^x_2 {ty(t)} \, dt]$         (Equation 2)

Using the second Fundamental Theorem of Calculus we know that:

$\frac{d}{dx} [\int\limits^x_2 {t} \, dt]=x\\ \frac{d}{dx} [\int\limits^x_2 {ty(t)} \, dt]=xy(x)$

So, we need to replace those equations in equation 2, and we obtain:

$\frac{dy}{dx} =x-xy(x)$       (Equation 3)

Now, we are going to resolve the equation 3 as a normal equation. So, we need to joint the same variables. I mean, variable y on one side and variable x on other side, as follows:

$\frac{dy}{dx} =x(1-y)\\\frac{dy}{(1-y)} =xdx\\\frac{dy}{y-1} =-xdx$

And, we integrate each side of the equation to obtain:

$ln|y-1|=-\frac{x^{2} }{2} +C$      (Equation 4)

Now, we need to find the value of the constant C. And we know that we can find one point of the equation, replacing x=2 in equation 1, because the integral becomes zero, so:

$y(2)=2+\int\limits^2_2 {t-ty(t)} \, dt=2$

And, we replace the value of y when x=2 in equation 4 and we obtain,

$ln|2-1|=-\frac{2^{2} }{2} +C\\C=2$

So, Equation 4 is:

$ln|y-1|=-\frac{x^{2} }{2} +2$         (Equation 4')

Now, we need to clear the y variable from Equation 4', (we are going to asumme that y-1>0),

$e^{ln(y-1)} =e^{-\frac{x^{2} }{2} +2}\\y-1=e^{-\frac{x^{2} }{2} +2}\\y=e^{-\frac{x^{2} }{2} +2}+1$