Question

Consider the initial value problem

Answer 1

Answer with Step-by-step explanation:

We are given that initial value problem

a.$4u''-u'+2u=0$

u(0)=3,u'(0)=0

Auxillary equation

$4m^2-m+2=0$

$m=\frac{1\pm\sqrt{(-1)^2-4(4)(2)}}{2(4)}$

By using quadratic formula:$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$m=\frac{1\pm\sqrt{1-32}}{8}$

$m=\frac{1\pm i\sqrt{31}}{8}=\frac{1}{8}\pm i\frac{\sqrt{31}}{8}$

Therefore, the general solution of the initial value problem

$u(t)=e^{\frac{t}{8}}(c_1cos\frac{\sqrt{31}}{8}t+c_2sin\frac{\sqrt{31}}{8}t)$

Substitute t=0 and u(0)=3

$3=c_1$

Differentiate w.r.t t

$u'(t)=\frac{1}{8}e^{\frac{t}{8}}(c_1cos\frac{\sqrt{31}}{8}t+c_2sin\frac{\sqrt{31}}{8}t)+e^{\frac{t}{8}}(-\frac{\sqrt{31}}{8}c_1sin\frac{\sqrt{31}}{8}t+\frac{\sqrt{31}}{8}c_2cos\frac{\sqrt{31}}{8}t)$

Substitute t=0 and u'(0)=0

$0=\frac{1}{8}c_1+\frac{\sqrt{31}}{8}c_2$

Substitute the value of $c_1$

$0=\frac{1}{8}\times 3+\frac{\sqrt{31}}{8}c_2$

$-\frac{3}{8}=\frac{\sqrt{31}}{8}c_2$

$c_2=-\frac{3}{8}\times \frac{8}{\sqrt{31}}$

$c_2=-\frac{3}{\sqrt{31}}$

Substitute the values

$u(t)=e^{\frac{t}{8}}(3cos\frac{\sqrt{31}}{8}t-\frac{3}{\sqrt{31}}sin\frac{\sqrt{31}}{8}t)$

(b) For t>0 $\mid u(t)\mid=10$

We cannot find the solution of  $\mid u(t)\mid=10$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  $\mid u(t)\mid=10$