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ipn [44]
2 years ago
14

Section 5.2 Problem 19:

Mathematics
1 answer:
rjkz [21]2 years ago
3 0

Answer:

y(x)=2xe^{3x} (See attached graph)

Step-by-step explanation:

To solve a second-order homogeneous differential equation, we need to substitute each term with the auxiliary equation am^2+bm+c=0 where the values of m are the roots:

y''-6y'+9y=0\\\\m^2-6m+9=0\\\\(m-3)^2=0\\\\m-3=0\\\\m=3

Since the values of m are equal real roots, then the general solution is y(x)=C_1e^{m_1x}+C_2xe^{m_1x}.

Thus, the general solution for our given differential equation is y(x)=C_1e^{3x}+C_2xe^{3x}.

To account for both initial conditions, take the derivative of y(x), thus, y'(x)=3C_1e^{3x}+C_2e^{3x}+3C_2xe^{3x}

Now, we can create our system of equations given our initial conditions:

y(x)=C_1e^{3x}+C_2xe^{3x}\\ \\y(0)=C_1e^{3(0)}+\frac{C_2}{6}(0)e^{3(0)}=0\\ \\C_1=0

y'(x)=3C_1e^{3x}+C_2e^{3x}+3C_2xe^{3x}\\\\y'(0)=3C_1e^{3(0)}+C_2e^{3(0)}+3C_2(0)e^{3(0)}=2\\\\3C_1+C_2=2

We then solve the system of equations, which becomes easy since we already know that C_1=0:

3C_1+C_2=2\\\\3(0)+C_2=2\\\\C_2=2

Thus, our final solution is:

y(x)=C_1e^{3x}+C_2xe^{3x}\\\\y(x)=2xe^{3x}

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Therefore, the function is continuous on the set of real numbers

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3.h(x)=\frac{3x-5}{x^2-5x+7}

x^2-5x+7=0

It cannot be factorize .

Therefore, it has no real values for which it is not defined .

Hence, function h is defined for all real values.

\lim_{x\rightarrow 5}\frac{3x-5}{x^2-5x+7}=\frac{15-5}{25-25+7}=\frac{10}{7}=1.43

2.g(x)=\frac{x-17}{x^2+75}

The denominator of g is defined for all real values of x.

Therefore, the function g is continuous on the set of real numbers

\lim_{x\rightarrow 5}\frac{x-17}{x^2+75}=\frac{5-17}{25+75}=\frac{-12}{100}=-0.12

4.i(x)=\frac{x^2-9}{x-9}

x-9=0

x=9

The function i is not defined for x=9

Therefore, the function i is  not continuous on the set of real numbers.

5.j(x)=\frac{4x^2-7x-65}{x^2+10}

The denominator of j is defined for all real values of x.

Therefore, the function j is continuous on the set of real numbers.

\lim_{x\rightarrow 5}\frac{4x^2-7x-65}{x^2+10}=\frac{100-35-65}{25+10}=0

6.k(x)=\frac{x+1}{x^2+x+29}

x^2+x+29=0

It cannot be factorize .

Therefore, it has no real values for which it is not defined .

Hence, function k is defined for all real values.

\lim_{x\rightarrow 5}\frac{x+1}{x^2+x+29}=\frac{5+1}{25+5+29}=\frac{6}{59}=0.102

7.l(x)=\frac{5x-1}{x^2-9x+8}

x^2-9x+8=0

x^2-8x-x+8=0

x(x-8)-1(x-8)=0

(x-8)(x-1)=0

x=8,1

The function is not defined for x=8 and x=1

Hence, function l is not  defined for all real values.

8.m(x)=\frac{x^2+5x-24}{x^2+11}

The denominator of m is defined for all real values of x.

Therefore, the function m is continuous on the set of real numbers.

\lim_{x\rightarrow 5}\frac{x^2+5x-24}{x^2+11}=\frac{25+25-24}{25+11}=\frac{26}{36}=\frac{13}{18}=0.722

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