Find the coordinates?

Question

Consider the initial value problem: y'+(1/2)y=2cos(t), y(0)=-1. Find the coordinates of the first local maximum point of the solution for t>0.

Answer 1

This is again a typical first-order linear differential equation: y' + Py = Q

Here is a standard way to solve it.

Multiply both sides of y' + Py = Q by e^∫P,

e^∫P y' + e^∫P Py = e^∫PQ

Apply product rule,

(e^∫P y)' = e^∫P Q

Solve for y,

y = (1/e^∫P)(∫e^∫P Q dx + c) <==Formula of the solution for any first-order linear differential equation

For your problem: y'+(1/2)y=2cos(t), y(0)=-1, and t>0

p = 1/2

e^∫P = e^(t/2)

y = e^(-t/2)[∫e^(t/2)2cos(t) dt + c] = (4/5)(2sin(t) + cos(t)) + ce(-t/2), integration by parts twice

Plug in y(0) = -1,

-1 = (4/5) + c => c = -9/5

y = (4/5)(2sin(t) + cos(t)) - (9/5)e^(-t/2)

y' = 0 => (4/5)[2cos(t) - sin(t)] + (9/10)e^(-t/2) = 0=> t = 1.364, where y' goes from positive to negative

The first local maximum point of the solution is at

t = 1.364, y(1.364) = 0.820 <==Answer

Answer 2

You can also follow the way used multiple times in early problems involving inhomogeneous differential equations: the general solution y₀ of the homogeneous equation (y' - (1/2)y = 0) plus a particular solution y₁of the inhomogeneous eqaution

                                                          y = y₀ + y₁

where y₀ = A e ^-t/2, and y₁ can be found in the form y₁ = a sin t   + b cos t. If you substitute this

expression into your original equation you will find

                                            a/2  - b = 0   and a + b/2 = 2

or   a = 8/5, b = 4/5. Thus, your solution is

                                          y = A e ^-t/2  + (8/5) sin t   +  (4/5) cos t

and A = -9/5 if you put y(0) = -1. Then, to find the maximum we have to put y' = 0 (y'>0 before this point and y'< 0 after)

                                          y' = (9/10) e ^-t/2  + (8/5) cos t - (4/5) sin t = 0

This equation can't be solved analytically. You have to apply numerical methods or have to use a graphing calculator, sketch the graph of the function y(t) and click on TRACE to find the point. 

According to my approximations y _max = 0.818 for t =1.367.

 

Answer 3

y'+(1/2)y=2cos(t)

vy' + 1/2 vy  =  2 v cos t                 Let  1/2  v  = dv/dt = v'

(vy)'  =   2 v cos t                                       1/2 dt   = dv/v

vy     =  2 int v cos t dt                             1/2 t  +  c   = ln v 

y  =  (1/v)* 2  int v cos t dt                   exp (1/2 t + c)   =  v   =   A  exp (t/2)

y   =  2*exp -(t/2)  int exp (t/2) cos t dt

 

Instead of integration by parts 2 times, let cos t  =   cos t + i sin t  = exp (it) 

After intergration we will take the Re part.

int exp (t/2) *exp (it) dt = int exp (t/2 + it) dt  = int exp t(1/2 + i) dt

= exp t(1/2 + i)/(1/2 + i)  =  (1/2 - i) exp t(1/2 + i)/(1/2 + i)(1/2 - i)                                                       

= (4/5)*(1/2 - i) exp t(1/2 + i)

=(4/5)*(1/2 - i) exp (t/2)*exp (it)

 

=(4/5)*(1/2 - i) exp (t/2)*(cos t + i sin t)

=(4/10) exp (t/2) (cos t + i sin t)  -  (4/5) exp (t/2) (i cos t - sin t), Take the real part of this.

= exp (t/2) ( 4/10 cos t   +  4/5 sin t)  + c

y = 2 exp -(t/2) *( exp (t/2) ( 4/10 cos t + 4/5 sin t)  +  c)

   =  4/5 cos t  + 8/5 sin t  + c exp(-t/2)

y(0) = -1 = 4/5 cos 0  +  8/5 sin 0  + c exp(-0/2)

-1  =  4/5  + c

c = - 9/5

the first local maximum occurs when y' = 0  =  -4/5 sin t  +  8/5 cos t  + 9/10 exp (-t/2)

 

Solve by iteration:

Using Octave, the code is:

function y = f (t);

y = (-4/5)*sin(t) + (8/5)*cos(t) + (9/10)*exp(-t/2);  #(Here y = y')#

endfunction;

#comment: Solve System for Root Starting at t0= 0

[t, info] = fsolve ("f", 0.)

 

t = 1.3643