The Solution.
The correct answer is option B.
Explanation:
This option is the only option with a constant change of 5 unit
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.
The answer to your question would be B.
I hope this helped:D
I believe that would be on the 15th day because the LCM, or least common multiple of 3 and 5 is 15.
Two distinct roots means two real solutions for x (the parabola needs to cross the x-axis twice)
Vertex form of a quadratic equation: (h,k) is vertex
y = a(x-h)^2 + k
The x of the vertex needs to equal 3
y = a(x-3)^2 + k
In order to have two distinct roots the parabola must be (+a) upward facing with vertex below the x-axis or (-a) downward facing with vertex above the x-axis. Parabolas are symmetrical so for an easy factorable equation make "a" 1 or -1 depending on if you want the upward/downward facing one.
y = (x-3)^2 - 1
Vertex (3,-1) upwards facing with two distinct roots 4 and 2
y = x^2 -6x + 9 - 1
y = x^2 -6x + 8
y = (x - 4)(x - 2)