Find the curve whose slope at the point (x, y) is 3x^2 if the curve has to pass through the point (1,-1)?
We know that the slope of a curve y = f(x) at the point (x, y) is given as dy/dx. So here, first we find the general solution of the differential equation by integrating both sides with respect to x:
=> y = Int [3x^2 dx]
= x^3 +C
Now to identify the particular solution whose graph passes through the point (1, -1), we find the value of C for which y= -1 when x = 1
y = x^3 +C
=> -1 = 1^3 +C
=> C = -1-1
=> C= -2
Therefore the solution we are looking for is y = x^3 – 2.
The slope of the curve at (x , y) =3x^2.
Therefore the curve must be the premtive of dy/dx = 3x^2.
So the curve is y = Integral (3x^2)dx
y = (3x^3)/4+ c, Or y = x^3+c ,where c is a constant. Since this curve passes through (1 , -1), the coordinates (1 , -1) should satisfy y = x^3+c.
-1 = 1^3+c.
c = -1-1 = -2
Therfore the required curve is y = x^3 -2.