Find y that satisfies the given condition
dy/dx=20x(5x^2-1)^3; curve passes through (1,3)
I took the antiderivative and ended up with 10x^2((5x^2-1)^4/4). I end up with a large number at the end, which I think might be wrong.
To find the value of y that satisfies the given condition, we need to integrate the differential equation:
dy/dx = 20x(5x^2 - 1)^3
To integrate, we can use the chain rule. Let u = 5x^2 - 1, then du/dx = 10x, and dy/dx = du/dx * dy/du. Rearranging this equation gives us dy/du * du/dx = 20x.
Now, we can separate the variables:
dy/du = 20x / du/dx
dy/du = 20x / 10x
dy/du = 2
Integrating both sides with respect to u gives us:
∫dy/du du = ∫2 du
y = 2u + C
Now we substitute u back in:
y = 2(5x^2 - 1) + C
y = 10x^2 - 2 + C
To find the value of C, we can use the given condition that the curve passes through the point (1,3). Plugging these values into the equation, we get:
3 = 10(1)^2 - 2 + C
3 = 10 - 2 + C
C = 3 - 8
C = -5
Therefore, the equation of the curve passing through (1,3) is:
y = 10x^2 - 2 - 5
y = 10x^2 - 7
So the value of y that satisfies the given condition is y = 10x^2 - 7.
let
u = 5x^2-1
du = 10x dx
so, what you have is 2u^3 du
y(x) = u^4/2 = 1/2 (5x^2-1)^4 + C
y(1) = 3, so
1/2 (5-1)^4+C = 3
2C = 6-256 = -250
y(x) = 1/2 (5x^2-1)^4 - 125
Had you checked your function, you would have seen that its derivative was not the original dy/dx.