find an equation of the curve that passes through the point (1,1) and whose slope at (x,y) is y^2/x^3
This means that:
dy/dx = y^2/x^3, and y(1)=1,
in the language of separation of variables.
Hint: use power rule for integration.
Also, see example to your previous question (under Billy):
http://www.jiskha.com/display.cgi?id=1300573436
Please, Catherine, Billy, Mark, Matthew, and Sally, use one single screen name if you want your posts to avoid confusion.
To find an equation of the curve that passes through the point (1,1) and whose slope at each point (x,y) is given by y^2/x^3, we can use the method of separation of variables.
Let's start by integrating both sides of the given equation with respect to x:
∫(1/y^2) dy = ∫(1/x^3) dx
Integrating the left side gives: -1/y = -1/(2x^2) + C1, where C1 is the constant of integration.
Multiplying both sides by -y, we get: 1 = (y/(2x^2)) - C1y.
Now, let's substitute the point (1,1) into the equation to solve for C1:
1 = (1/(2(1)^2)) - C1(1)
1 = 1/2 - C1
C1 = 1/2
Substituting C1 back into the equation, we have:
1 = (y/(2x^2)) - (1/2)y
Multiplying both sides by 2x^2 gives:
2x^2 = y - (1/2)y
2x^2 = (1/2)y
4x^2 = y
Therefore, the equation of the curve that passes through the point (1,1) and has the slope y^2/x^3 is y = 4x^2.
So, the equation of the curve is y = 4x^2.