Answer:
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Step-by-step explanation:
Answer:
The answer to your math eqaution is 7/8
Answer:
Step-by-step explanation:
16x^2 + 9 = 25
16x^2 = 16
x^2 = 1
x = 1, -1
x*y' + y = 8x
y' + y/x = 8 .... divide everything by x
dy/dx + y/x = 8
dy/dx + (1/x)*y = 8
We have something in the form
y' + P(x)*y = Q(x)
which is a first order ODE
The integrating factor is 
Multiply both sides by the integrating factor (x) and we get the following:
dy/dx + (1/x)*y = 8
x*dy/dx + x*(1/x)*y = x*8
x*dy/dx + y = 8x
y + x*dy/dx = 8x
Note the left hand side is the result of using the product rule on xy. We technically didn't need the integrating factor since we already had the original equation in this format, but I wanted to use it anyway (since other ODE problems may not be as simple).
Since (xy)' turns into y + x*dy/dx, and vice versa, this means
y + x*dy/dx = 8x turns into (xy)' = 8x
Integrating both sides with respect to x leads to
xy = 4x^2 + C
y = (4x^2 + C)/x
y = (4x^2)/x + C/x
y = 4x + Cx^(-1)
where C is a constant. In this case, C = -5 leads to a solution
y = 4x - 5x^(-1)
you can check this answer by deriving both sides with respect to x
dy/dx = 4 + 5x^(-2)
Then plugging this along with y = 4x - 5x^(-1) into the ODE given, and you should find it satisfies that equation.
Answer:
Step-by-step explanation:
Given that,
y' = 17y ( 1-y^7)
Let y=1
Then, y' = 0 for all t
Then show that it is the only stable equilibrium point so that as y→1, t→∞ with any initial value.
So, the graph solution will be
y(0) = 1 and this will be an horizontal line
If, y(0) > 1 then, y' < 0 by inspecting the first equation, so the graph is has decreasing solution.
Likewise, if y(0) < 1 then, y' > 0 and the graph is increasing.
So no matter the initial condition, graph of the solution will be asymptotic to the horizontal line above.
This make the limit be 1.
This shows that x = 1 is a stable equilibrium.