For #1 - 12 determine if the equation is separable, linear, or neither. If the function is separable or linear find a general solution. If you can make this solution explicit, then do so.
\(y' = yt\)
\(ty' = y+1\)
\(t\dot{y}-y^2\sin(t) = e^t\)
\(y' = \frac{2t}{y^4+4y^3}\)
\(y\dot{y} +\frac{\log(t)}{t} = 1\).
\(\dot{\theta} = \frac{\theta+t}{\theta+1}\)
\(x' + \frac{x^2}{\tan(t)} =0\)
\(t\dot{z} = e^z\log(t)\)
\(w'wt + \sin(tw) = 0\)
\(x^2y' + xy = x^3\)
\(\frac{dy}{dx}=\frac{x(e^{x^2}+1)}{3y^2}\)
\[y'-3y=ty\] Solve the above differential equation using two different methods.
For #13- 22 find solutions to the given initial value problem. If you can find an explicit solution then do so.
\( y' + y^2\cos(t) = 0\) where \(y(0) = \frac{1}{2}.\)
\(\frac{1}{t^2} + \dfrac{y'}{\sqrt{y}}=0; \, \, y(-1)=1.\)
\(x' = e^{t+x}\) where \(x(0)=0\)
\(y' = \frac{\cos(t)}{5y^4+2y}\) where \(y(0) = 2\)
\(u' = \frac{u^2+1}{t^2-1}\) where \(u(0) = 1\)
\(e^ty' = \frac{y}{e^{-t}-1}\) where \(y(1) = 1\)
\(ty' = y^3\) where \(y(1) = y_0\). Your solution will depend on the value \(y_0\), be sure you indicate how!
\(y' = \frac{t}{y(1-t^2)}; \quad y(0) = y_0\) Your solution will depend on the value \(y_0\), be sure you indicate how!
\(y' = \frac{2t}{\log(y)}\) where \(y(0) = y_0\). Your solution will depend on the value \(y_0\), be sure you indicate how!
\(t\theta\theta'=1 \) where \(\theta(1) = \theta_0\) Your solution will depend on the value \(\theta_0\), be sure you indicate how!
In the following justify your responses.
\(v'=\frac{(v-1)^{\frac{1}{3}}}{1+t^2} \) where \(v(0) = v_0\) Your solution will depend on the value \(y_0\), be sure you indicate how! Then indicate which value of \(v_0\) makes the solution not unique.
The goal of this problem is to analyze separable problems in a similar way to how we analyzed explicit equations. Recall from chapter 2 that an equation is explicit if it is of the form \(y' = f(t)\). We solved such equations by applying the Fundamental Theorem of Calculus to integrate both sides and get an explicit solution for \(y\).
Now consider the separable equation \(\frac{dy}{dt} = \frac{f(t)}{g(y)}\) so that the equation separates as \(g(y)dy = f(t)dt\).
(a) Suppose \(G(y) +C_1 = \int g(y)dy\). Use implicit differentiation to find the derivative of \(G(y) \) with respect to \(t\).
(b) Suppose \(F(t) +C_2 = \int f(t)dt\). What does the FTC tell you the derivative of \(F(t)\) with respect to \(t\) is?
(c) Use (a), (b) and the FTC to justify why \(G(y) = F(t) +C\) is an implicit general solution to the given separable equation.
Suppose we deposit \(\$1000\) into a savings account with \(5\%\) interest compounded continuously. After \(5\) years we start taking \(\$200\) out each year. How long will the money last?
What parameter values will guarantee that \(x(t)\) will go to infinity as \(t \rightarrow \infty\)? \[\frac{dx}{dt}=\alpha x(t-\beta) \,\]