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3 years ago

PLEASE HELP LAST PROBLEM I NEED HELP ON (Homework Help)

Mathematics

1 answer:

V125BC [204]3 years ago

7 0

Answer:

Step-by-step explanation:

so the 2 in front means stretch 0f 2 the -3 in the parenthaesis means right 3 and the 1 means up 1

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Evaluate. (15+12)2+310⋅5 Enter your answer as a mixed number in simplest form by filling in the boxes. $$

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Hey there!

(15+12)(2)+ 310(5)

= 27(2) + 310(5)

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3 years ago

Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 Preview

mario62 [17]

Answer: The required solution is

$y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.$

Step-by-step explanation: We are given to solve the following differential equation :

$5y^{\prime\prime}+3y^\prime-2y=0,~~~~~~~y(0)=0,~~y^\prime(0)=2.8~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$y=e^{mt}$ be an auxiliary solution of equation (i).

Then, we have

$y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.$

Substituting these values in equation (i), we get

$5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.$

So, the general solution of the given equation is

$y(t)=Ae^{-t}+Be^{\frac{1}{5}t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-Ae^{-t}+\dfrac{B}{5}e^{\frac{1}{5}t}.$

According to the given conditions, we have

$y(0)=0\\\\\Rightarrow A+B=0\\\\\Rightarrow B=-A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

and

$y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.$

From equation (ii), we get

$B=\dfrac{7}{3}.$

Thus, the required solution is

$y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.$

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3 years ago