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; 5y³/3 = 2t - 15t²/2 + 5/3;

y³(1) = (2 - 15/2 + 5/3)·3/5 = - 11/2·3/5 +1 = - 33/10 + 1 = - 2.3000