Respuesta :
Answer:
Step-by-step explanation:
Given that the solution of a certain differential equation is of the form
[tex]y(t) = ae^{7t} +be^{11t}[/tex]
Use the initial conditions
i) y(0) =1
[tex]1=a(1)+b(1)\\a+b=1[/tex] ... I
ii) y'(0) = 4
Find derivative of y first and then substitute
[tex]y'(t) = 7ae^{7t} +11be^{11t}\\y'(0) =7a+11b \\7a+11b =4 ...II[/tex]
Now using I and II we solve for a and b
Substitute b = 1-a in II
[tex]7a+11(1-a) = 4\\-4a+11 =4\\-4a =-7\\a = 1.75 \\b = -0.75[/tex]
Hence solution is
[tex]y(t) = 1.75e^{7t} -0.75e^{11t}[/tex]
Answer:
y(t) = a exp(3t) + b exp(4t) conditions, y(0) = 3 y'(0) = 3 y(0) = a exp(3 x 0) + b exp(4 x 0) = a exp(0) + b exp(0) = (a x 1) + (b x 1) = a + b y'(0) = 0 so the linear equation is, a + b = 3
Step-by-step explanation:
