Question

Consider the ordinary differential equation (answer questions in picture)

Answer 1

a. Given the 2nd order ODE

$y''(x) = 4y(x) + 4$

if we substitute $z(x)=y'(x)+2y(x)$ and its derivative, $z'(x)=y''(x)+2y'(x)$, we can eliminate $y(x)$ and $y''(x)$ to end up with the ODE

$z'(x) - 2y'(x) = 4\left(\dfrac{z(x)-y'(x)}2\right) + 4$

$z'(x) - 2y'(x) = 2z(x) - 2y'(x) + 4$

$\boxed{z'(x) = 2z(x) + 4}$

and since $y(0)=y'(0)=1$, it follows that $z(0)=y'(0)+2y(0)=3$.

b. I'll solve with an integrating factor.

$z'(x) = 2z(x) + 4$

$z'(x) - 2z(x) = 4$

$e^{-2x} z'(x) - 2 e^{-2x} z(x) = 4e^{-2x}$

$\left(e^{-2x} z(x)\right)' = 4e^{-2x}$

$e^{-2x} z(x) = -2e^{-2x} + C$

$z(x) = -2 + Ce^{2x}$

Since $z(0)=3$, we find

$3 = -2 + Ce^0 \implies C=5$

so the particular solution for $z(x)$ is

$\boxed{z(x) = 5e^{-2x} - 2}$

c. Knowing $z(x)$, we recover a 1st order ODE for $y(x)$,

$z(x) = y'(x) + 2y(x) \implies y'(x) + 2y(x) = 5e^{-2x} - 2$

Using an integrating factor again, we have

$e^{2x} y'(x) + 2e^{2x} y(x) = 5 - 2e^{2x}$

$\left(e^{2x} y(x)\right)' = 5 - 2e^{2x}$

$e^{2x} y(x) = 5x - e^{2x} + C$

$y(x) = 5xe^{-2x} - 1 + Ce^{-2x}$

Since $y(0)=1$, we find

$1 = 0 - 1 + Ce^0 \implies C=2$

so that

$\boxed{y(x) = (5x+2)e^{-2x} - 1}$

Answer 2

6.a) The differential equation for z(x) is z'(x) = 2z(x) + 4, z(0) = 3.

6.b) The value of z(x) is $z(x) = 5e^{2x} - 2$.

6.c) The value of y(x) is $y(x) = \frac{5e^{2x}}{4} - \frac{1}{4e^{2x}} -1$.

The given ordinary differential equation is y''(x) = 4y(x) + 4, y(0) = y'(0) = 1 ... (d).

We are also given a substitution function, z(x) = y'(x) + 2y(x) ... (z).

Putting x = 0, we get:

z(0) = y'(0) + 2y(0),

or, z(0) = 1 + 2*1 = 3.

Rearranging (z), we can write it as:

z(x) = y'(x) + 2y(x),

or, y'(x) = z(x) - 2y(x) ... (i).

Differentiating (z) with respect to x, we get:

z'(x) = y''(x) + 2y'(x),

or, y''(x) = z'(x) - 2y'(x) ... (ii).

Substituting the value of y''(x) from (ii) in (d) we get:

y''(x) = 4y(x) + 4,

or, z'(x) - 2y'(x) = 4y(x) + 4.

Substituting the value of y'(x) from (i) we get:

z'(x) - 2y'(x) = 4y(x) + 4,

or, z'(x) - 2(z(x) - 2y(x)) = 4y(x) + 4,

or, z'(x) - 2z(x) + 4y(x) = 4y(x) + 4,

or, z'(x) = 2z(x) + 4y(x) - 4y(x) + 4,

or, z'(x) = 2z(x) + 4.

The initial value of z(0) was calculated to be 3.

6.a) The differential equation for z(x) is z'(x) = 2z(x) + 4, z(0) = 3.

Transforming z(x) = dz/dx, and z = z(x), we get:

dz/dx = 2z + 4,

or, dz/(2z + 4) = dx.

Integrating both sides, we get:

∫dz/(2z + 4) = ∫dx,

or, {ln (z + 2)}/2 = x + C,

or, $\sqrt{z+2} = e^{x + C}$,

or, $z =Ce^{2x}-2$ ... (iii).

Substituting z = 3, and x = 0,  we get:

$3 = Ce^{2*0} - 2\\\Rightarrow C - 2 = 3\\\Rightarrow C = 5.$

Substituting C = 5, in (iii), we get:

$z = 5e^{2x} - 2$.

6.b) The value of z(x) is $z(x) = 5e^{2x} - 2$.

Substituting the value of z(x) in (z), we get:

z(x) = y'(x) + 2y(x),

or, 5e²ˣ - 2 = y'(x) + 2y(x),

which gives us:

$y(x) = \frac{5e^{2x}}{4} - \frac{1}{4e^{2x}} -1$ for the initial condition y(x) = 0.

6.c) The value of y(x) is $y(x) = \frac{5e^{2x}}{4} - \frac{1}{4e^{2x}} -1$.

Learn more about differential equations at

https://brainly.com/question/18760518

#SPJ1