Back to the class
Quiz 3 solution
Solve $y'=(t^2-4)(2y-3)$.
Solution: First rewrite using $\dfrac{\mathrm{d}y}{\mathrm{d}t}$: $$\dfrac{\mathrm{d}y}{\mathrm{d}t} = (t^2-4)(2y-3).$$ Now separate variables and integrate to get $$\displaystyle\int \dfrac{1}{2y-3} \mathrm{d}y = \displaystyle\int t^2-4 \mathrm{d}t.$$ Compute the integrals to arrive at $$\dfrac{1}{2}\ln(|2y-3|) = \dfrac{t^3}{3}-4t+C.$$ Multiply by $2$ to get $$\ln(|2y-3|) = \dfrac{2t^3}{3}-8t+2C.$$ Apply the exponenital function to both sides to get $$|2y-3| = e^{\frac{2t^3}{3}}e^{-8t}e^{2C}.$$ Remove the absolute value wih $\pm$ (easiest to put it with the constant $C$): $$2y-3 = \pm e^{2C} e^{\frac{2t^3}{3}} e^{-8t}.$$ Now solve for $y$ to get the general solution (we put all constant factors together into $\tilde{C}$!) $$y(t)=\dfrac{3}{2} + \tilde{C}e^{\frac{2t^3}{3}}e^{-8t}, \quad \tilde{C}=\dfrac{\pm e^{2C}}{2}.$$