$\frac{a}{4} + 6 = \frac{a}{8} + 8$
you will multiply by 8 now
$2a + 48 = a + 64$
now move a's with a's and the numbers that don't have a variable with each other
$2a - a = 64 - 48$
now subtract
$a = 16$
I hope it helps you!

4 0

3 years ago

Snezhnost [94]

6a. By the convolution theorem,

$L\{t^3\star e^{5t}\} = L\{t^3\} \times L\{e^{5t}\} = \dfrac6{s^4} \times \dfrac1{s-5} = \boxed{\dfrac5{s^4(s-5)}}$

6b. Similarly,

$L\{e^{3t}\star \cos(t)\} = L\{e^{3t}\} \times L\{\cos(t)\} = \dfrac1{s-3} \times \dfrac s{1+s^2} = \boxed{\dfrac s{(s-3)(s^2+1)}}$

7. Take the Laplace transform of both sides, noting that the integral is the convolution of $e^t$ and $f(t)$ .

$\displaystyle f(t) = 3 - 4 \int_0^t e^\tau f(t - \tau) \, d\tau$

$\implies \displaystyle F(s) = \dfrac3s - 4 F(s) G(s)$

where $g(t) = e^t$ . Then $G(s) = \frac1{s-1}$ , and

$F(s) = \dfrac3s - \dfrac4{s-1} F(s) \implies F(s) = \dfrac{\frac3s}{\frac{s+3}{s-1}} = 3\dfrac{s-1}{s(s+3)}$

We have the partial fraction decomposition,

$\dfrac{s-1}{s(s+3)} = \dfrac13 \left(-\dfrac1s + \dfrac4{s+3}\right)$

Then we can easily compute the inverse transform to solve for f(t) :

$F(s) = -\dfrac1s + \dfrac4{s+3}$

$\implies \boxed{f(t) = -1 + 4e^{-3t}}$

6 0

1 year ago

Marina CMI [18]

Answer:

Move all terms that don't contain y to the right side and solve.y=4−6x/5

Step-by-step explanation:

4 0

2 years ago