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4 years ago

PLEASE HELP! can anyone who answers this provide explanations and how to solve each and as well? ;-;

Mathematics

1 answer:

sergeinik [125]4 years ago

8 0

Answer:

1) 6.32

2)8.66

3)5.74

4)9.49

5)16.55

6)21.21

7)25.98

Step-by-step explanation:

1) 6,7

2) 8,9

3) 5,6

4) 9,10

5) 16,17

6) 21,22

7) 25,26

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Solve this problem by using variation of parameters method.<br> y''-y=coshx.

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For the nonhomogeneous ODE

$y''-y=\underbrace{\cosh x}_{f(x)}$

we're looking for a particular solution of the form

$y_p=u_1y_1+u_2y_2$

where

$u_1=-\displaystyle\int\frac{y_2(x)f(x)}{W(y_1(x),y_2(x))}\,\mathrm dx$
$u_2=\displaystyle\int\frac{y_1(x)f(x)}{W(y_1(x),y_2(x))}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the two fundamental solutions.

We have

$W(y_1,y_2)=\begin{vmatrix}\cosh x&\sinh x\\\sinh x&\cosh x\end{vmatrix}=\cosh^2x-\sinh^2x=1$

so we're left with

$u_1=-\displaystyle\int\sinh x\cosh x\,\mathrm dx=-\dfrac12\cosh^2x$
$u_2=\displaystyle\int\cosh^2x\,\mathrm dx=\dfrac12x+\dfrac14\sinh2x$

so that the particular solution is

$y_p=-\dfrac12\cosh^3x+\dfrac12x\sinh x+\dfrac14\sinh x\sinh2x$
$y_p=-\dfrac12\cosh x+\dfrac12x\sinh x$

As $y_1$ already accounts for the $\cosh x$ term in $y_p$ , we're left with the general solution

$y=C_1\cosh x+C_2\sinh x+\dfrac12x\sinh x$

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