Answer with Step-by-step explanation:
We are given that initial value problem
a.[tex]4u''-u'+2u=0[/tex]
u(0)=3,u'(0)=0
Auxillary equation
[tex]4m^2-m+2=0[/tex]
[tex]m=\frac{1\pm\sqrt{(-1)^2-4(4)(2)}}{2(4)}[/tex]
By using quadratic formula:[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
[tex]m=\frac{1\pm\sqrt{1-32}}{8}[/tex]
[tex]m=\frac{1\pm i\sqrt{31}}{8}=\frac{1}{8}\pm i\frac{\sqrt{31}}{8}[/tex]
Therefore, the general solution of the initial value problem
[tex]u(t)=e^{\frac{t}{8}}(c_1cos\frac{\sqrt{31}}{8}t+c_2sin\frac{\sqrt{31}}{8}t)[/tex]
Substitute t=0 and u(0)=3
[tex]3=c_1[/tex]
Differentiate w.r.t t
[tex]u'(t)=\frac{1}{8}e^{\frac{t}{8}}(c_1cos\frac{\sqrt{31}}{8}t+c_2sin\frac{\sqrt{31}}{8}t)+e^{\frac{t}{8}}(-\frac{\sqrt{31}}{8}c_1sin\frac{\sqrt{31}}{8}t+\frac{\sqrt{31}}{8}c_2cos\frac{\sqrt{31}}{8}t)[/tex]
Substitute t=0 and u'(0)=0
[tex]0=\frac{1}{8}c_1+\frac{\sqrt{31}}{8}c_2[/tex]
Substitute the value of [tex]c_1[/tex]
[tex]0=\frac{1}{8}\times 3+\frac{\sqrt{31}}{8}c_2[/tex]
[tex]-\frac{3}{8}=\frac{\sqrt{31}}{8}c_2[/tex]
[tex]c_2=-\frac{3}{8}\times \frac{8}{\sqrt{31}}[/tex]
[tex]c_2=-\frac{3}{\sqrt{31}}[/tex]
Substitute the values
[tex]u(t)=e^{\frac{t}{8}}(3cos\frac{\sqrt{31}}{8}t-\frac{3}{\sqrt{31}}sin\frac{\sqrt{31}}{8}t)[/tex]
(b) For t>0 [tex]\mid u(t)\mid=10[/tex]
We cannot find the solution of [tex]\mid u(t)\mid=10[/tex]analytically
But we can use the graph to find the approximate value of the solution.
By graph , the approximate value of t=12.2 for which [tex]\mid u(t)\mid=10[/tex]
