dy/dx = 2y^2
Integrating...y=2/3 y^3 + C
put 1,-1 into the equation, and solve for C.
Then find the y for x=2
if y= a^uhttp://math2.org/math/integrals/tableof.htm
see exponential functions.
dy/dx=2y^2 and if y=-1 when x=1, then when x=2, y=?
how do i get x in the equation. do i integrate dy/dx because that would be (2/3)y^3 +C.
2)how i would integrate 13^x-11^x.
lets say i integrate 13^x, would that be 13^x*log 13. i know there is a formula for that but i don't remember
To find the value of y when x=2 for the equation dy/dx = 2y^2, you first integrate both sides of the equation with respect to x.
Integrating dy/dx gives you ∫dy = ∫2y^2 dx. This simplifies to y = ∫2y^2 dx.
Now, integrating the right side requires knowledge of the integral of y^2. To integrate y^2, use the power rule for integration: ∫y^n dy = (1/(n+1)) * y^(n+1). Applying this to the integral of 2y^2 gives (2/(2+1)) * y^(2+1) = (2/3) * y^3 + C, where C is the constant of integration.
To complete the solution, you need to find the value of C. Use the given information that y=-1 when x=1. Substituting these values into the equation y = (2/3) * y^3 + C gives -1 = (2/3) * (-1)^3 + C.
Solving this equation for C, you get C = -1 - (2/3). Simplifying, C = -5/3.
Now that you have the value of C, you can find the value of y when x=2. Substituting x=2 and the value of C into the equation y = (2/3) * y^3 + C gives y = (2/3) * y^3 - 5/3. To solve for y, you can use numerical techniques or approximation methods as the equation is non-linear.