Consider the equation \(\dfrac{\dy}{\dt} = 2\sin(t)\cos(t)\).
(a) Show that \(y(t) = \sin^2(t) \) is a solution to this equation for every \(t\) in \((-\infty,\infty)\).
(b) Give a general solutions to this equation.
Show that \(y_1(t) = \sqrt{t^2 -1}\) and \(y_2(t) = - \sqrt{t^2 - 9}\) are both solutions to \[\frac{\dy}{\dt} = \frac{t}{y} \quad \text{for every $t$ in $(3,\infty)$} \,.\]
In #3-7 find a general solutions to the given equation.
\(y' = 8x^3 + 10x\)
\(\dot{y} = \dfrac{1}{t^2}\)
\(\dfrac{\dw}{\dt} = \sec^2(t)\)
\(\dot{x} = 6e^{-3t}\)
\(y'' = 2t\) (Note: This is a second order equation, so you will have to integrate twice, and your general solution for \(y\) will depend on two distinct constants).
In #8- 12 find the largest interval over which the initial-value problem has a unique solution and find the solution on that interval.
\(x' = \log|t| \,, \quad x(1) = 2 \,.\)
\(y' = \frac{1}{t^2-2t} \,, \quad y(1) = 1 \,.\)
\(\sin(t) y' = \cos(t) \,, \quad y(\frac{\pi}{2}) = 1 \,.\)
\(y' = 4e^{2x} \,, \quad y(0) = 1 \,.\)
\(h'' = -9.8 \,, \qquad h'(0) = v_0 \,, \quad h(0) = h_0 \,.\) Note: this is a second-order explicit equation, but it is a first-order explicit equation in \(h'\), and using the first initial condition we can get a first order equation in \(h\).
In the following justify your responses.
\(y' = \dfrac{-1}{\sqrt{1-t^2}} \,, \quad y(\frac{1}{2}) = \frac{2\pi}{3} \,.\)
(a) We know that given an explicit first order equation \(y' = f(t)\) we can apply the Fundamental Theorem of Calculus and get a general solution is given by the indefinite integral \(y(t) = \int f(t)\,\dt\). Justify why the solution to the initial-value problem \(y' = f(t) \,, \quad y(t_0) = y_0\) is given by this expression with a definite integral: \[y(t) = y_0 + \int_{t_0}^t f(u) \du \,.\] (b) In a sentence describe how you would find the largest interval where the solution found in part(a) to the initial value problem is unique.
Every initial value problem can be transformed so that the initial condition is at the origin. This can be obtained via the equivalence: \(y'=f(t,y)\) with initial condition \(y(t_0) = y_0\) is equivalent to \(w' = f(s, w)\) if we set \(s=t-t_0\) and \(w=y-y_0\) we have the initial equivalent initial condition is \(w(0) = 0\). \(y(t)\) is a solution to the first IVP iff \(w(s)\) is a solution to the second IVP.
Rewrite each of the following IVP as their equivalent IVP centered at the origin:
(a) \(y' = 5 t \,, \quad y(1) = 2 \,.\)
(b) \(y' = y + 1 \,, \quad y(2) = -1 \,.\)
(c) \(y' = y t \,, \quad y(-1) = 1 \,.\)
Find a general solution of \(y' = \dfrac{1}{(x-\alpha)(x-\beta)}\,,\) where \(\alpha\neq\beta\) .
Confirm that \(y(t)=1+t\) is a solution to \(\dfrac{\dy}{\dt}=\dfrac{y^2-1}{t^2+2t}\,.\)