In each of Problems 1 through 12: (a) Draw a direction field for the given differential equation.
(b) Based on an inspection of the direction field, describe how solutions behave for large $t$.
(c) Find the general solution of the given differential equation, and use it to determine how solutions behave as $t\to \infty$.
1) $y' + 3y = t + e^{-2t}$
6) $ty' + 2y = sin (t)$, $t>0$
10) $ty' - y = t^2e{-t}$, $t > 0$
In each of Problems 13 through 20 find the solution of the given initial value problem.
13) $y' - y = 2te^{2t}$, $y(0) = 1$
17) $y' - 2y = e^{2t}$, $y(0) = 2$
18) $ty' + 2y = \sin (t)$, $y(\pi/2) = 1$, $t > 0$
19) $t^3y' + 4t^2y = e^{-t}$, $y(-1) = 0$, $t < 0$
28) Consider the initial value problem $$ y' + \frac{2}{3} y = 1 - \frac{1}{2} t,\qquad y(0) = y_0. $$ Find the value of $y_0$ for which the solution touches, but does not cross, the $t$-axis.
30) Find the value of $y_0$ for which the solution of the initial value problem $$ y' - y = 1 + 3 \sin (t),\qquad y(0) = y_0 $$ remains finite as $t\to \infty$.
38) Variation of Parameters. Consider the following method of solving the general linear equation of first order: $$ y' + p(t)y = g(t). \tag{i} $$ (a) If $g(t) = 0$ for all $t$, show that the solution is $$ y = A\exp\bigl[-\int p(t)\,dt \bigr] \tag{ii} $$ where $A$ is a constant.
(b) If $g(t)$ is not everywhere zero, assume that the solution of Eq. (i) is of the form $$ y = A(t)\exp\bigl[-\int p(t)\,dt \bigr] \tag{iii} $$ where $A$ is now a function of $t$. By substituting for $y$ in the given differential equation, show that $A(t)$ must satisfy the condition $$ A'(t) = \bigl[\int p(t)\,dt \bigr] \tag{iv} $$ (c) Find $A(t)$ from Eq. (iv). Then substitute for A(t) in Eq. (iii) and determine $y$. Verify that the solution obtained in this manner agrees with that of Eq. (33) in the text. This technique is known as the method of variation of parameters; it is discussed in detail in Section 3.6 in connection with second order linear equations.
In each of Problems 39 through 42 use the method of Problem 38 to solve the given differential equation
40) $y' + (1/t)y = 3 \cos (2t)$, $t > 0$
41) $ty' + 2y = \sin (t)$, $t > 0$
In each of Problems 1 through 8 solve the given differential equation.
3) $y' + y^2 \sin (x) =0$
6) $xy' = (1 - y^2)^{1/2}$
7) $\displaystyle{\frac{dy}{dx}= \frac{x - e^{-x}}{y + e^y}}$
In each of Problems 9 through 20:
(a) Find the solution of the given initial value problem in explicit form.
(b) Plot the graph of the solution.
(c) Determine (at least approximately) the interval in which the solution is defined.
15) $y' = 2x/(1 + 2y)$, $y(2) = 0$
20) $y^2(1 - x^2)^{1/2}\,dy = \arcsin (x)\,dx$, $y(0) = 1$
26) Solve the initial value problem $$ y' = 2(1 + x)(1 + y^2),\qquad y(0) = 0 $$ and determine where the solution attains its minimum value.
30) Consider the equation $$ \frac{dy}{dx}= \frac{y - 4x}{x - y}. \tag{i} $$ (a) Show that Eq. (i) can be rewritten as $$ \frac{dy}{dx}= \frac{(y/x) - 4}{1 - (y/x)}.\tag{ii} $$ thus Eq. (i) is homogeneous.
(b) Introduce a new dependent variable $v$ so that $v = y/x$, or $y = xv(x)$. Express $dy/dx$ in terms of $x, v$, and $dv/dx$.
(c) Replace $y$ and $dy/dx$ in Eq. (ii) by the expressions from part (b) that involve $v$ and $dv/dx$. Show that the resulting differential equation is $$ v + x\frac{dv}{dx}=\frac{v - 4}{1 - v}, $$ or $$ x\frac{dv}{dx}= \frac{v^2 - 4}{1 - v}.\tag{iii} $$ Observe that Eq. (iii) is separable.
(d) Solve Eq. (iii), obtaining $v$ implicitly in terms of $x$.
(e) Find the solution of Eq. (i) by replacing $v$ by $y/x$ in the solution in part (d).
(f) Draw a direction field and some integral curves for Eq. (i). Recall that the right side of Eq. (i) actually depends only on the ratio $y/x$. This means that integral curves have the same slope at all points on any given straight line through the origin, although the slope changes from one line to another. Therefore the direction field and the integral curves are symmetric with respect to the origin. Is this symmetry property evident from your plot? The method outlined in Problem 30 can be used for any homogeneous equation. That is, the substitution $y = xv(x)$ transforms a homogeneous equation into a separable equation. The latter equation can be solved by direct integration, and then replacing $v$ by $y/x$ gives the solution to the original equation. In each of Problems 31 through 38:
(a) Show that the given equation is homogeneous.
(b) Solve the differential equation.
(c) Draw a direction field and some integral curves. Are they symmetric with respect to the origin?
35) $\displaystyle{\frac{dy}{dx}= \frac{x + 3y}{x - y}}$