A first order linear equation in the form y′+p(x)y=f(x) can be solved by finding an integrating factor μ(x)=exp(∫p(x)dx) (1) Given the equation (x+2)2y′+5(x+2)y=16 find μ(x)= (2) Then an explicit general solution with arbitrary constant C can be written in the form (x+2)5y= +C. (3) Then solve the initial value problem with y(0)=4 y= .

The question is not well-typed, but I understand that:

A first order linear equation in the form

y′ + p(x)y = f(x) (1)

can be solved by finding an integrating factor

μ(x) = e^(∫p(x)dx) (2)

(1) Given the equation

(x + 2)²y′ + 5(x + 2)y = 16 (3)

find μ(x).

SOLUTION:

First, we need the equation (3) to be in the form of (1). To do this, we will divide the equation by (x + 2)². So we have

y′ + 5/(x + 2)y = 16/(x + 2)²

This is now in the form of (1), with

p(x) = 5/(x + 2), and f(x) = 16/(x + 2)² .

The integrating factor can now be obtained using (2).

μ(x) = e^{∫[5/(x + 2)]dx}

= e^[5ln(x + 2)]

μ(x) = (x + 2)^5

(2) Then an explicit general solution with arbitrary constant C can be written in the form ((x + 2)^5)y= f(x) + C.

SOLUTION

To obtain an explicit general solution of (3), first, we multiply (3) by the integrating factor. Doing that, we have:

((x + 2)^5)y′ + 5((x + 2)^4)y = 16(x + 2)³

This can be expressed as

d(y(x + 2)^5) = 16(x + 2)³

Integrating both sides, we have

y(x + 2)^5 = 4(x + 2)^4 + C(4)

Which is the form we are required to have.

(3) Then solve the initial value problem with y(0) = 4

SOLUTION

Given the condition y(0) = 4, we put y = 4, and x = 0 in the general solution (4) obtained to obtain the value of the arbitrary constant C.

Doing that, we have

4(0 + 2)^5 = 4(0 + 2)^4 + C

C = 4(2^5) - 4(2^4)

= 128 - 64 = 64

Finally, we insert C = 64 in (4) to have the required particular solution. Doing that, we have

y(x + 2)^5 = 4(x + 2)^4 + 64

And that is it.

no, because the sides are not congruent

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.