Respuesta :
Answer:
[tex]y(x)=e^{\frac{-x^{2}}{2}+2 }+1[/tex]
Step-by-step explanation:
We have that:
[tex]y(x)=2+\int\limits^x_2 {[t-ty(t)]} \, dt[/tex] (Equation 1)
To resolve this integral equation, we need to use the second Fundamental Theorem of Calculus, which says:
[tex]\frac{d}{dx} [\int\limits^x_a {f(t)} \, dt]=f(x)[/tex]
So, we need to differentiate both sides of equation 1 with respect to x:
[tex]\frac{dy}{dx} =\frac{d}{dx} [2+\int\limits^x_2 {[t-ty(t)]} \, dt][/tex]
[tex]\frac{dy}{dx} =\frac{d(2)}{dx}+\frac{d}{dx} [\int\limits^x_2 {[t-ty(t)]} \, dt][/tex]
We know that the derivate for a constant value is zero. And,
[tex]\frac{dy}{dx}=\frac{d}{dx} [\int\limits^x_2 {t} \, dt]-\frac{d}{dx}[\int\limits^x_2 {ty(t)} \, dt][/tex] (Equation 2)
Using the second Fundamental Theorem of Calculus we know that:
[tex]\frac{d}{dx} [\int\limits^x_2 {t} \, dt]=x\\ \frac{d}{dx} [\int\limits^x_2 {ty(t)} \, dt]=xy(x)[/tex]
So, we need to replace those equations in equation 2, and we obtain:
[tex]\frac{dy}{dx} =x-xy(x)[/tex] (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:
[tex]\frac{dy}{dx} =x(1-y)\\\frac{dy}{(1-y)} =xdx\\\frac{dy}{y-1} =-xdx[/tex]
And, we integrate each side of the equation to obtain:
[tex]ln|y-1|=-\frac{x^{2} }{2} +C[/tex] (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:
[tex]y(2)=2+\int\limits^2_2 {t-ty(t)} \, dt=2[/tex]
And, we replace the value of y when x=2 in equation 4 and we obtain,
[tex]ln|2-1|=-\frac{2^{2} }{2} +C\\C=2[/tex]
So, Equation 4 is:
[tex]ln|y-1|=-\frac{x^{2} }{2} +2[/tex] (Equation 4')
Now, we need to clear the y variable from Equation 4', (we are going to asumme that y-1>0),
[tex]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[/tex]
