Solve the given initial value problem.

Question

5y^2(dy/dt)+15t=2, y(0)=1

Let y(t) be the solution of the initial value problem. Find the value of y^3(1) with the answer rounded to 4 digits after the decimal point.

Answer 1

5y²dy = (2 - 15t)dt; ∫5y²dy = ∫(2 - 15t)dt; 5y³/3 = 2t - 15t²/2 + C;5/3 = C; so, 5y³/3 = 2t - 15t²/2 + 5/3

5/3y³(1) = 2 - 15/2 + 5/3; y³(1) = - 23/10; = - 2.3