Respuesta :
Answer:
[tex]y_0 = -1.643[/tex]
Step-by-step explanation:
Assuming the following function:
[tex] y' + \frac{2}{3}y = 1-\frac{1}{2}t , y(0)=y_0[/tex]
So we want to find a point a such that:
[tex] y(a) = 0 , y'(a) =0 , y''(a) \neq 0[/tex]
So if we use this condition we can find the value of t=a that satisfy the equirements:
[tex] 0 +\frac{2}{3} 0 = 1- \frac{1}{2}a[/tex]
[tex] a=2[/tex]
And we have another initial conidtion [tex]y(2)=0, y'(2)=0[/tex]
Now we need to solve the differential equation and since is a linear equation we can use the integrating factor, our equation have the following form:
[tex]y'(x) + p(x) y = q(x)[/tex]
[tex] p(x) = \frac{2}{3}, q(x) = 1-\frac{t}{2}[/tex]
We can calculate the integrating factor like this:
[tex] \mu(t) = e^{\int \frac{2}{3}dt} =e^{\frac{2t}{3}}[/tex]
Now we can rewrite the differential equation like this:
[tex] [e^{\frac{2t}{3}}y]' = e^{\frac{2t}{3}}-\frac{e^{2t/3}}{2}t[/tex]
And if we integrate both sides we got this:
[tex] ye^{\frac{2t}{3}}= \frac{3}{2}e^{\frac{2t}{3}}-\frac{1}{2} [\frac{3}{2}te^{\frac{2t}{3}}-\frac{9}{4}e^{\frac{2t}{3}}]+C[/tex]
If we divide both sides by [tex]e^{\frac{2t}{3}} [/tex] we got this:
[tex] y = \frac{21}{8} -\frac{3}{4} t +Ce^{-\frac{2t}{3}} [/tex]
And we can use the initial condition [tex]y(0)=y_0[/tex] and we find the value for C like this:
[tex] C = y_0 - \frac{21}{8}[/tex]
And then we have our solution given by:
[tex] y =\frac{21}{8} -\frac{3}{4}t +(y_0 -\frac{21}{8})e^{-\frac{2t}{3}}[/tex]
And if we use the other initial condition [tex] y(2) =0[/tex] w can solve the value of [tex]y_0[/tex]
[tex]0 =\frac{21}{8} -\frac{3}{4}2 +(y_0 -\frac{21}{8})e^{-\frac{4}{3}}[/tex]
[tex] y_0 = \frac{\frac{21}{8} (e^{-\frac{4}{3}} -1)+ \frac{3}{2}}{e^{-\frac{4}{3}}}=-1.643[/tex]
And then that would be our final solution:
[tex]y_0 = -1.643[/tex]
