Question

Solve the given initial-value problem in which the input function g(x) is discontinuous. [Hint: Solve the problem on two intervals, and then find a solution so that y and y' are continuous at x = π/2.] y'' + 4y = g(x), y(0) = 1, y'(0) = 6, where g(x) = sin(x), 0 ≤ x ≤ π/2 0, x > π/2

Answer 1

Step-by-step explanation:

$g(x)=y''+4y\\\\y''+4y=0\\\\m^2+4=0\\m_{1,2} =+2i,-2i$

so the complimentary solution is

$y_{c} =c_{1} cos2x+c_{2} sin2x$

solving for particular solution we get

$y_{p} =acosx+bsinx$

to find the values of constants a,b we put values in main equation and

$-acosx-bsinx+4acosx+4bsinx=sinx$

as g(x)=sin x

so above equation yields us

$a=0,b=1/3$

so writing the solution

$y=y_{c} +y_{p} \\y=c_{1} cos2x+c_{2} sin2x+(1/3)sinx$

using IC's to get the values for $c_{1} ,c_{2}$

$c_{1}=1 ,c_{2} =5/6\\so\\y(x)=cos2x+(5/6)sin2x+(1/3)sinx ,0\leq x\leq \frac{\pi }{2}$

solving for upper range

$y(x)=\frac{2}{3} cos2x+\frac{5}{6} sin2x,x>\frac{\pi }{2}$

Answer 2

The exercise above is one of Discontinuous Initial Value Problem (DIVP) with a Discontinuous Function.

What is an Initial value problem?

This topic is related to multivariable calculus. It is a standard differential equation with an initial condition that describes the unknown function's value at a selected point in the domain. IVP is often used in physics for modeling.

What is the solution for the IVP whose input function g(x) is discontinuous?

y" + 4y = g(x),

y(0) =1, and

y' (0) = 2

Where

g(x) = $\left[\begin{array}{cc}sin x,&0\leq x\leq \pi /2\4&0,\7&x > \pi /2\end{array}\right]$

Next, we solve for the lower interval and use the result to solve the upper interval.

Homogeneous solution to (1) where (1) is

y′′ + 4y = g(x), y(0) = 1, y′(0) = 2

We have:

m² + 4 = 0 → m₁,₂ = ± 2i

From this we derive:

yh = c₁ Cos 2x + c₂ sin 2x

To resolve the particular solution, we state that:

yp = a cosx + b sin x

Resolving the constants a and b, let's substitute back into equation 1 above to arrive at:

-a cos x - b sin x + 4a cos x + 4b sin x = sinx

From the above, we have:

a = 0 and b= 1/3

Rewriting the equation, we have:

y = yh + yp = c₁ cos 2x + c₂ sin 2x + 1/3sinx

Using the Initiation constraints, we solve for c₁ and c₂:

c₁ = 1, c₂ = 5/6, this leaves us with the following:


y(x) =cos2x + 5/6sin 2x + 1/3sinx,

0$\leq x\leq \pi /2$.........................................(2)

From the above, we must proceed to calculate the upper half of the range so that we have, y" + 4y = 0.

Using the initial constraints in combination with (2), we get y $(\pi /2)$ = -(2/3) and

y' ($\pi$/2) = -(5/3)

Recall that solution 2 was:

y = c₁ cos 2x + c₂ sin 2x

With the new constraints, we can resolve this to get:

c₁ = 2/3, c₂ = 5/6, so

y(x) = 2/3cos 2x + 5/6sin 2x, x > $\pi$/3...............(3)

Combining equations 2 and 3, we have:

y(x) = $\left[\begin{array}{cc}cos 2x + 5/6sing2x +1/3sinx,&0\leq x\leq \pi /2\4&2/3cos2x +5/6sin2x,\7&x > \pi /2\end{array}\right]$

Learn more about discontinuous IVP at:

https://brainly.com/question/14986286