Find the function y that satisfies the following conditions dy/dx = 2x+2/ 3√x ;y(1)=2. y=?

Question

Find the function y that satisfies the following conditions

dy/dx = 2x+2/ 3√x ;y(1)=2.

y=?

Answer 1

First, one must find the integral of the function. Assuming the dependent variable is y and the independent variable is x, one must solve the equation for y:

1. dy/dx = 2x + [ 2 / (3 √x) ]
2. dy = [ 2x + (2/3)x^-1/2 ] dx
3. ∫ dy = ∫ [ 2x + (2/3)x^-1/2 ] dx
4. ∫ dy = ∫ 2x dx + ∫ (2/3)x^-1/2 dx
5. ∫ dy = 2 ∫ x dx + (2/3) ∫ x^-1/2 dx
6. y + C₁ = (2/2)x² + (2/3)(2)x^1/2 + C₂
7. y = x² + (4/3)x^1/2 + C; where C = C₂ - C₁

The equation is:

y = x² + (4/3) √x + C

Now, the value of the constant C must be determined. The problem states y(1) = 2; in other words, when x = 1, y = 2. Using these two values in the last equation and solving for C:

1. x² + (4/3) √x + C = y
2. (1)² + (4/3) √1 + C = 2
3. 1 + 4/3 + C = 2
4. 7/3 + C = 2
5. C = 2 - 7/3
6. C = -1/3

Substituting this value for C in the last equation one obtains the final form of the function:

1. y = x² + (4/3) √x + C
2. y = x² + (4/3) √x - 1/3

Final function is:

y = x² + (4/3) √x - 1/3

Answer 2

y = ∫dy = ∫2x +(2/3)x^-1/2dx = x²+(2/3)(2)x^1/2 + C

y(1) = 2 = 1 + 4/3 + C --> C = -1/3

y = x²+(4/3)√x -1/3