Respuesta :
Answer:
[tex]y=\frac{-7t^2+22t-7}{7t-22}[/tex]
Step-by-step explanation:
We are given that
Initial value problem
[tex]y'=(t+y)^2-1[/tex], y(3)=4
Substitute the value [tex]z=t+y[/tex]
When t=3 and y=4 then
z=3+4=7
[tex]y'=z^2-1[/tex]
Differentiate z w.r.t t
Then, we get
[tex]\frac{dz}{dt}=1+y'[/tex]
[tex]z'=1+z^2-1=z^2[/tex]
[tex]z^{-2}dz=dt[/tex]
Integrate on both sides
[tex]-\frac{1}{z}dz=t+C[/tex]
[tex]z=-\frac{1}{t+C}[/tex]
Substitute t=3 and z=7
Then, we get
[tex]7=-\frac{1}{3+C}[/tex]
[tex]21+7C=-1[/tex]
[tex]7C=-1-21=-22[/tex]
[tex]C=-\frac{22}{7}[/tex]
Substitute the value of C then we get
[tex]z=-\frac{1}{t-\frac{22}{7}}[/tex]
[tex]z=\frac{-7}{7t-22}[/tex]
[tex]y=z-t[/tex]
[tex]y=\frac{-7}{7t-22}-t[/tex]
[tex]y=\frac{-7-7t^2+22t}{7t-22}[/tex]
[tex]y=\frac{-7t^2+22t-7}{7t-22}[/tex]
The differential equation rewritten as a separable equation of the form z′=g(z)y is; [8/(4 - 3e^(2t - 6))] - 1 - t
How to rewrite differential equations?
We are given;
y' = (t + y)² - 1
Put u = t + y such that u' = y'
u' = u² - 1 which is inseparable and as such;
u'/(u² - 1) = 1
We Integrate both sides with respect to t;
(u'/2)[(1/(u - 1)) - (1/(u + 1))] = 1
∫(u'/2)[(1/(u - 1)) - (1/(u + 1))] dt = ∫dt
⇒ ¹/₂(In |u - 1| - In |u + 1|) = t + C
¹/₂In (|u - 1|/|u + 1|) = t + C
In|1 - (2/(u + 1))| = 2t + C
1 - (2/(u + 1)) = Ce^(2t)
(2/(u + 1)) = 1 - Ce^(2t)
u = [2/(1 - Ce^(2t))] - 1
Putting back t + y for u gives;
y = [2/(1 - Ce^(2t))] - 1 - t
At initial condition; y(3) = 4, Thus;
4 = [2/(1 - Ce^(2*3))] - 1 - 3
Solving for C gives; C = 3/4e⁶
Thus, the particular solution is;
y = [8/(4 - 3e^(2t - 6))] - 1 - t
Read more about differential equations at; https://brainly.com/question/18760518
