2.2 Separable Equations

A first order differential equation is separable if it can be written as

\begin{equation} h(y)y'=g(x), \tag{2.2.1}\end{equation}

where the left side is a product of \(y'\) and a function of \(y\) and the right side is a function of \(x\). Rewriting a separable differential equation in this form is called separation of variables. In Section 2.1 we used separation of variables to solve homogeneous linear equations. In this section we’ll apply this method to nonlinear equations.

To see how to solve (2.2.1), let’s first assume that \(y\) is a solution. Let \(G(x)\) and \(H(y)\) be antiderivatives of \(g(x)\) and \(h(y)\); that is,

\begin{equation} H'(y)=h(y)\mbox{\quad and \quad}G'(x)=g(x). \tag{2.2.2}\end{equation}

Then, from the chain rule,

\[ {d\over dx}H(y(x))=H'(y(x))y'(x)=h(y)y'(x). \]

Therefore (2.2.1) is equivalent to

\[ {d\over dx}H(y(x))={d\over dx}G(x). \]

Integrating both sides of this equation and combining the constants of integration yields

\begin{equation} H(y(x))=G(x)+c. \tag{2.2.3}\end{equation}

Although we derived this equation on the assumption that \(y\) is a solution of (2.2.1), we can now view it differently: Any differentiable function \(y\) that satisfies (2.2.3) for some constant \(c\) is a solution of (2.2.1). To see this, we differentiate both sides of (2.2.3), using the chain rule on the left, to obtain

\[ H'(y(x))y'(x)=G'(x), \]

which is equivalent to

\[ h(y(x))y'(x)=g(x) \]

because of (2.2.2).

In conclusion, to solve (2.2.1) it suffices to find functions \(G=G(x)\) and \(H=H(y)\) that satisfy (2.2.2). Then any differentiable function \(y=y(x)\) that satisfies (2.2.3) is a solution of (2.2.1).

Example 2.2.1

Solve the equation

\[ y'=x(1+y^2). \]

Solution Separating variables yields

\[ {y'\over 1+y^2}=x. \]

Integrating yields

\[ \tan^{-1}y={x^2\over2}+c \]

Therefore

\[ y=\tan\left({x^2\over2}+c\right). \]

Example 2.2.2

Solve the equation

\begin{equation} y'=-{x\over y}. \tag{2.2.4}\end{equation}
Solve the initial value problem

\begin{equation} y'=-{x\over y}, \quad y(1)=1. \tag{2.2.5}\end{equation}
Solve the initial value problem

\begin{equation} y'=-{x\over y}, \quad y(1)=-2. \tag{2.2.6}\end{equation}

Solution (a) Separating variables in (2.2.4) yields

\[ yy'=-x. \]

Integrating yields

\[ {y^2\over2}=-{x^2\over2}+c,\mbox{\quad or,\, equivalently, \quad} x^2+y^2=2c. \]

The last equation shows that \(c\) must be positive if \(y\) is to be a solution of (2.2.4) on an open interval. Therefore we let \(2c=a^2\) (with \(a > 0\)) and rewrite the last equation as

\begin{equation} x^2+y^2=a^2. \tag{2.2.7}\end{equation}

This equation has two differentiable solutions for \(y\) in terms of \(x\):

\begin{equation} y=\phantom{-} \sqrt{a^2-x^2}, \quad -a < x < a, \tag{2.2.8}\end{equation}

and

\begin{equation} y= - \sqrt{a^2-x^2}, \quad -a < x < a. \tag{2.2.9}\end{equation}

The solution curves defined by (2.2.8) are semicircles above the \(x\)-axis and those defined by (2.2.9) are semicircles below the \(x\)-axis (Figure 2.2.1).

Solution (b) The solution of (2.2.5) is positive when \(x=1\); hence, it is of the form (2.2.8). Substituting \(x=1\) and \(y=1\) into (2.2.7) to satisfy the initial condition yields \(a^2=2\); hence, the solution of (2.2.5) is

\[ y=\sqrt{2-x^2}, \quad - \sqrt{2}< x < \sqrt{2}. \]

Solution (c) The solution of (2.2.6) is negative when \(x=1\) and is therefore of the form (2.2.9). Substituting \(x=1\) and \(y=-2\) into (2.2.7) to satisfy the initial condition yields \(a^2=5\). Hence, the solution of (2.2.6) is

\[ y=- \sqrt{5-x^2}, \quad -\sqrt{5} < x < \sqrt{5}. \]

(a) y=2-x^2,\, -2<x<2; \, (b) y=-5-x^2,\, -5<x<5 — Figure 2.2.1. **(a)** \(y=\sqrt{2-x^{2}}\), \(-\sqrt{2}<x<\sqrt{2}\); **(b)** \(y=-\sqrt{5-x^{2}}\), \(-\sqrt{5}<x<\sqrt{5}\)

Implicit Solutions of Separable Equations

In Examples 2.2.1 and 2.2.2 we were able to solve the equation \(H(y)=G(x)+c\) to obtain explicit formulas for solutions of the given separable differential equations. As we’ll see in the next example, this isn’t always possible. In this situation we must broaden our definition of a solution of a separable equation. The next theorem provides the basis for this modification. We omit the proof, which requires a result from advanced calculus called as the implicit function theorem.

Theorem 2.2.1

Suppose \(g=g(x)\) is continous on \((a,b)\) and \(h=h(y)\) are continuous on \((c,d).\) Let \(G\) be an antiderivative of \(g\) on \((a,b)\) and let \(H\) be an antiderivative of \(h\) on \((c,d).\) Let \(x_0\) be an arbitrary point in \((a,b),\) let \(y_0\) be a point in \((c,d)\) such that \(h(y_0)\ne0,\) and define

\begin{equation} c=H(y_0)-G(x_0). \tag{2.2.10}\end{equation}

Then there’s a function \(y=y(x)\) defined on some open interval \((a_1,b_1),\) where \(a\le a_1<x_0<b_1\le b,\) such that \(y(x_0)=y_0\) and

\begin{equation} H(y)=G(x)+c \tag{2.2.11}\end{equation}

for \(a_1<x<b_1\). Therefore \(y\) is a solution of the initial value problem

\begin{equation} h(y)y'=g(x),\quad y(x_0)=x_0. \tag{2.2.12}\end{equation}

It’s convenient to say that (2.2.11) with \(c\) arbitrary is an implicit solution of \(h(y)y'=g(x)\). Curves defined by (2.2.11) are integral curves of \(h(y)y'=g(x)\). If \(c\) satisfies (2.2.10), we’ll say that (2.2.11) is an implicit solution of the initial value problem (2.2.12). However, keep these points in mind:

For some choices of \(c\) there may not be any differentiable functions \(y\) that satisfy (2.2.11).
The function \(y\) in (2.2.11) (not (2.2.11) itself) is a solution of \(h(y)y'=g(x)\).

Example 2.2.3

Find implicit solutions of

\begin{equation} y'={2x+1\over5y^4+1}. \tag{2.2.13}\end{equation}
Find an implicit solution of

\begin{equation} y'={2x+1\over5y^4+1},\quad y(2)=1. \tag{2.2.14}\end{equation}

Solution (a) Separating variables yields

\[ (5y^4+1)y'=2x+1. \]

Integrating yields the implicit solution

\begin{equation} y^5+y=x^2+x+ c. \tag{2.2.15}\end{equation}

of (2.2.13).

Solution (b) Imposing the initial condition \(y(2)=1\) in (2.2.15) yields \(1+1=4+2+c\), so \(c=-4\). Therefore

\[ y^5+y=x^2+x-4 \]

is an implicit solution of the initial value problem (2.2.14). Although more than one differentiable function \(y=y(x)\) satisfies 2.2.13) near \(x=1\), it can be shown that there’s only one such function that satisfies the initial condition \(y(1)=2\).

Figure 2.2.2 shows a direction field and some integral curves for (2.2.13).

A direction field and integral curves for y'=2x+15y^4+1 — Figure 2.2.2. A direction field and integral curves for \(y'=\dst{\frac{2x+1}{5y^{4}+1}}\)

Constant Solutions of Separable Equations

An equation of the form

\[ y'=g(x)p(y) \]

is separable, since it can be rewritten as

\[ {1\over p(y)}y'=g(x). \]

However, the division by \(p(y)\) is not legitimate if \(p(y)=0\) for some values of \(y\). The next two examples show how to deal with this problem.

Example 2.2.4

Find all solutions of

\begin{equation} y'=2xy^2. \tag{2.2.16}\end{equation}

Solution Here we must divide by \(p(y)=y^2\) to separate variables. This isn’t legitimate if \(y\) is a solution of (2.2.16) that equals zero for some value of \(x\). One such solution can be found by inspection: \(y \equiv 0\). Now suppose \(y\) is a solution of (2.2.16) that isn’t identically zero. Since \(y\) is continuous there must be an interval on which \(y\) is never zero. Since division by \(y^2\) is legitimate for \(x\) in this interval, we can separate variables in (2.2.16) to obtain

\[ {y'\over y^2}=2x. \]

Integrating this yields

\[ -{1\over y}=x^2+c, \]

which is equivalent to

\begin{equation} y=-{1\over x^2+c}. \tag{2.2.17}\end{equation}

We’ve now shown that if \(y\) is a solution of (2.2.16) that is not identically zero, then \(y\) must be of the form (2.2.17). By substituting (2.2.17) into (2.2.16), you can verify that (2.2.17) is a solution of (2.2.16). Thus, solutions of (2.2.16) are \(y\equiv0\) and the functions of the form (2.2.17). Note that the solution \(y\equiv0\) isn’t of the form (2.2.17) for any value of \(c\).

Figure 2.2.3 shows a direction field and some integral curves for (2.2.16)

A direction field and integral curves for y'=2xy^2 — Figure 2.2.3. A direction field and integral curves for \(y'=2xy^{2}\)

Example 2.2.5

Find all solutions of

\begin{equation} y'={1\over2}x(1-y^2). \tag{2.2.18}\end{equation}

Solution Here we must divide by \(p(y)=1-y^2\) to separate variables. This isn’t legitimate if \(y\) is a solution of (2.2.18) that equals \(\pm1\) for some value of \(x\). Two such solutions can be found by inspection: \(y \equiv 1\) and \(y\equiv-1\). Now suppose \(y\) is a solution of (2.2.18) such that \(1-y^2\) isn’t identically zero. Since \(1-y^2\) is continuous there must be an interval on which \(1-y^2\) is never zero. Since division by \(1-y^2\) is legitimate for \(x\) in this interval, we can separate variables in (2.2.18) to obtain

\[ {2y'\over y^2-1}=-x. \]

A partial fraction expansion on the left yields

\[ \left[{1\over y-1}-{1\over y+1}\right]y'=-x, \]

and integrating yields

\[ \ln\left|{y-1\over y+1}\right|=-{x^2\over2}+k; \]

hence,

\[ \left|{y-1\over y+1}\right|=e^ke^{-x^2/2}. \]

Since \(y(x)\ne\pm1\) for \(x\) on the interval under discussion, the quantity \((y-1)/(y+1)\) can’t change sign in this interval. Therefore we can rewrite the last equation as

\[ {y-1\over y+1}=ce^{-x^2/2}, \]

where \(c=\pm e^k\), depending upon the sign of \((y-1)/(y+1)\) on the interval. Solving for \(y\) yields

\begin{equation} y={1+ce^{-x^2/2}\over 1-ce^{-x^2/2}}. \tag{2.2.19}\end{equation}

We’ve now shown that if \(y\) is a solution of (2.2.18) that is not identically equal to \(\pm1\), then \(y\) must be as in (2.2.19). By substituting (2.2.19) into (2.2.18) you can verify that (2.2.19) is a solution of (2.2.18). Thus, the solutions of (2.2.18) are \(y\equiv1\), \(y\equiv-1\) and the functions of the form (2.2.19). Note that the constant solution \(y \equiv 1\) can be obtained from this formula by taking \(c=0\); however, the other constant solution, \(y \equiv -1\), can’t be obtained in this way.

Figure 2.2.4 shows a direction field and some integrals for (2.2.18).

A direction field and integral curves for y'=x(1-y^2)2 — Figure 2.2.4. A direction field and integral curves for \(y'=\dst{\frac{x(1-y^2)}{2}}\)

Differences Between Linear and Nonlinear Equations

Theorem 2.1.2 states that if \(p\) and \(f\) are continuous on \((a,b)\) then every solution of

\[ y'+p(x)y=f(x) \]

on \((a,b)\) can be obtained by choosing a value for the constant \(c\) in the general solution, and if \(x_0\) is any point in \((a,b)\) and \(y_0\) is arbitrary, then the initial value problem

\[ y'+p(x)y=f(x),\quad y(x_0)=y_0 \]

has a solution on \((a,b)\).

The not true for nonlinear equations. First, we saw in Examples 2.2.4 and 2.2.5 that a nonlinear equation may have solutions that can’t be obtained by choosing a specific value of a constant appearing in a one-parameter family of solutions. Second, it is in general impossible to determine the interval of validity of a solution to an initial value problem for a nonlinear equation by simply examining the equation, since the interval of validity may depend on the initial condition. For instance, in Example 2.2.2 we saw that the solution of

\[ {dy\over dx}=-{x\over y},\quad y(x_0)=y_0 \]

is valid on \((-a,a)\), where \(a=\sqrt{x_0^2+y_0^2}\).

Example 2.2.6

Solve the initial value problem

\[ y'=2xy^2, \quad y(0)=y_0 \]

and determine the interval of validity of the solution.

Solution First suppose \(y_0\ne0\). From Example 2.2.4, we know that \(y\) must be of the form

\begin{equation} y=-{1\over x^2+c}. \tag{2.2.20}\end{equation}

Imposing the initial condition shows that \(c=-1/y_0\). Substituting this into (2.2.20) and rearranging terms yields the solution

\[ y= {y_0\over 1-y_0x^2}. \]

This is also the solution if \(y_0=0\). If \(y_0<0\), the denominator isn’t zero for any value of \(x\), so the the solution is valid on \((-\infty,\infty)\). If \(y_0>0\), the solution is valid only on \((-1/\sqrt{y_0},1/\sqrt{y_0})\).

2.2 Exercises

In Exercises 1–6 find all solutions.

\(\dst{y'={3x^2+2x+1\over y-2}}\)

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\(y=2\pm\sqrt{2(x^3+x^2+x+c)}\)
\((\sin x)(\sin y)+(\cos y)y'=0\)

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\(\ln (|\sin y|) = \cos x+c\); \(y \equiv k \pi\), \(k=\) integer
\(xy'+y^2+y=0\)

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\(\dst{y={c\over x-c} \quad y \equiv -1}\)
\(y' \ln |y|+x^2y= 0\)

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\(\dst{(\ln y)^2\over2}=-{x^3\over3}+c\)
\(\dst{(3y^3+3y \cos y+1)y'+{(2x+1)y\over 1+x^2}=0}\)

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\(y^3+3 \sin y + \ln |y|+\ln (1+x^2)+\tan^{-1}x=c\); \(y\equiv0\)
\(x^2yy'=(y^2-1)^{3/2}\)

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\(y=\pm\dst{\left(1+\left(x\over1+cx\right)^2\right)^{1/2}}\); \(y\equiv\pm1\)

In Exercises 7–10 find all solutions. Also, plot a direction field and some integral curves on the indicated rectangular region.

C/G \(\dst{y'=x^2(1+y^2)}; \; \{-1\le x\le1,\ -1\le y\le1\}\)

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\(\dst{y=\tan \left( {x^3\over3}+c\right)}\)
C/G \( y'(1+x^2)+xy=0 ; \; \{-2\le x\le2,\ -1\le y\le1\}\)

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\(y=\dst{c\over\sqrt{1+x^2}}\)
C/G \(y'=(x-1)(y-1)(y-2); \; \{-2\le x\le2,\ -3\le y\le3\}\)

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\(\dst{y= {2 - ce^{(x-1)^2/2}\over 1-ce^{(x-1)^2/2}}; \quad y \equiv 1}\)
C/G \((y-1)^2y'=2x+3; \; \{-2\le x\le2,\ -2\le y\le5\}\)

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\(\dst{y=1+\big(3x^2+9x+c)^{1/3}}\)

In Exercises 11 and 12 solve the initial value problem.

\(\dst{y'={x^2+3x+2\over y-2}, \quad y(1)=4}\)

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\(\dst{y=2+\sqrt{{2\over3}x^3+3x^2+4x -{11\over3}}}\)
\(y'+x(y^2+y)=0, \quad y(2)=1\)

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\(\dst{y={e^{-(x^2-4)/2} \over2-e^{-(x^2-4)/2}}}\)

In Exercises 13-16 solve the initial value problem and graph the solution.

C/G \((3y^2+4y)y'+2x+\cos x=0, \quad y(0)=1\)

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\(y^3+2y^2+x^2+\sin x=3\)
C/G \(\dst{y'+{(y+1)(y-1)(y-2)\over x+1}=0, \quad y(1)=0}\)

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\(\dst{(y+1)(y-1)^{-3}(y-2)^2=-256(x+1)^{-6}}\)
C/G \(y'+2x(y+1)=0, \quad y(0)=2\)

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\(y=-1+3e^{-x^2}\)
C/G \(y'=2xy(1+y^2),\quad y(0)=1\)

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\(y=\dst{1\over\sqrt{2e^{-2x^2}-1}}\)

In Exercises 17–23 solve the initial value problem and find the interval of validity of the solution.

\(y'(x^2+2)+ 4x(y^2+2y+1)=0, \quad y(1)=-1\)

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\(y \equiv -1; \quad (- \infty,\infty)\)
\(y'=-2x(y^2-3y+2), \quad y(0)=3\)

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\(\dst{y={4-e^{-x^2}\over2-e^{-x^2}}; \quad (-\infty,\infty)}\)
\(\dst{y'={2x\over 1+2y}, \quad y(2)=0}\)

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\(\dst{y={-1+\sqrt{4x^2-15} \over2}; \quad\left({\sqrt{15}\over2}, \infty\right)}\)
\(y'=2y-y^2, \quad y(0)=1\)

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\(y=\dst{\frac{2}{1+e^{-2x}}}\) \((-\infty,\infty)\)
\(x+yy'=0, \quad y(3) =-4\)

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\(y=- \sqrt{25-x^2}\); \((-5,5)\)
\(y'+x^2(y+1)(y-2)^2=0, \quad y(4)=2\)

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\(y \equiv 2, \quad (- \infty,\infty)\)
\((x+1)(x-2)y'+y=0, \quad y(1)=-3\)

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\(y=3\dst{\left(x+1\over2x-4\right)^{1/3}}\); \((-\infty,2)\)
Solve \(\dst{y'={(1+y^2) \over (1+x^2)}}\) explicitly. Hint: Use the identity \(\dst{\tan(A+B)={\tan A+\tan B\over1-\tan A\tan B}}\).

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\(y=\dst{x+c\over1-cx}\)
Solve \(\dst {y'\sqrt{1-x^2}+\sqrt{1-y^2}=0}\) explicitly. Hint: Use the identity \(\sin(A-B)=\sin A\cos B-\cos A\sin B\).

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\(y=-x\cos c+\sqrt{1-x^2}\sin c; \quad y \equiv 1; y \equiv -1\)
Solve \(\dst{y'={\cos x\over \sin y},\quad y (\pi)={\pi\over2}}\) explicitly. Hint: Use the identity \(\cos(x+\pi/2)=-\sin x\) and the periodicity of the cosine.

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\(y=-x+3\pi/2\)
Solve the initial value problem

\[ y'=ay-by^2,\quad y(0)=y_0. \]

Discuss the behavior of the solution if (a) \(y_0\ge0\); (b) \(y_0<0\).
The population \(P=P(t)\) of a species satisfies the logistic equation

\[ P'=aP(1-\alpha P) \]

and \(P(0)=P_0>0\). Find \(P\) for \(t>0\), and find \(\lim_{t\to\infty}P(t)\).

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\(P=\dst{P_0\over\alpha P_0+(1-\alpha P_0)e^{-at}}\); \(\lim_{t\to\infty}P(t)=1/\alpha\)
An epidemic spreads through a population at a rate proportional to the product of the number of people already infected and the number of people susceptible, but not yet infected. Therefore, if \(S\) denotes the total population of susceptible people and \(I=I(t)\) denotes the number of infected people at time \(t\), then

\[ I'=rI(S-I), \]

where \(r\) is a positive constant. Assuming that \(I(0)=I_0\), find \(I(t)\) for \(t>0\), and show that \(\lim_{t\to\infty}I(t)=S\).

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\(I=\dst{SI_0\over I_0+(S-I_0)e^{-rSt}}\)
L The result of Exercise 29 is discouraging: if any susceptible member of the group is initially infected, then in the long run all susceptible members are infected! On a more hopeful note, suppose the disease spreads according to the model of Exercise 29, but there’s a medication that cures the infected population at a rate proportional to the number of infected individuals. Now the equation for the number of infected individuals becomes

\[ I'=rI(S-I)-qI \tag*{\rm(A)} \]

where \(q\) is a positive constant.
1. Choose \(r\) and \(S\) positive. By plotting direction fields and solutions of (A) on suitable rectangular grids
  
  \[ R=\{0\le t \le T,\ 0\le I \le d\} \]
  
  in the \((t,I)\)-plane, verify that if \(I\) is any solution of (A) such that \(I(0)>0\), then \(\lim_{t\to\infty}I(t)=S-q/r\) if \(q<rS\) and \(\lim_{t\to\infty}I(t)=0\) if \(q\ge rS\).
2. To verify the experimental results of (a), use separation of variables to solve (A) with initial condition \(I(0)=I_0>0\), and find \(\lim_{t\to\infty}I(t)\). Hint: There are three cases to consider: (i) \(q<rS\); (ii) \(q>rS\); (iii) \(q=rS\).
Show answer

If \(q=rS\) then \(I=\dst{I_0\over1+rI_0t}\) and \(\lim_{t\to\infty}I(t)=0\). If \(q\ne Rs\), then \(I=\dst{\alpha I_0\over I_0+(\alpha-I_0)e^{-r\alpha t}}\). If \(q<rs\), then \(\lim_{t\to\infty}I(t)=\alpha=S-\dst{q\over r}\)

if \(q>rS\), then \(\lim_{t\to\infty}I(t)=0\)
L Consider the differential equation

\[ y'=ay-by^2-q, \tag*{\rm(A)} \]

where \(a\), \(b\) are positive constants, and \(q\) is an arbitrary constant. Suppose \(y\) denotes a solution of this equation that satisfies the initial condition \(y(0)=y_0\).
1. Choose \(a\) and \(b\) positive and \(q<a^2/4b\). By plotting direction fields and solutions of (A) on suitable rectangular grids
  
  \[ R=\{0\le t \le T,\ c\le y \le d\} \tag*{\rm(B)} \]
  
  in the \((t,y)\)-plane, discover that there are numbers \(y_1\) and \(y_2\) with \(y_1<y_2\) such that if \(y_0>y_1\) then \(\lim_{t\to\infty}y(t)=y_2\), and if \(y_0<y_1\) then \(y(t)=-\infty\) for some finite value of \(t\). (What happens if \(y_0=y_1\)?)
2. Choose \(a\) and \(b\) positive and \(q=a^2/4b\). By plotting direction fields and solutions of (A) on suitable rectangular grids of the form (B), discover that there’s a number \(y_1\) such that if \(y_0\ge y_1\) then \(\lim_{t\to\infty}y(t)=y_1\), while if \(y_0<y_1\) then \(y(t)=-\infty\) for some finite value of \(t\).
3. Choose positive \(a\), \(b\) and \(q>a^2/4b\). By plotting direction fields and solutions of (A) on suitable rectangular grids of the form (B), discover that no matter what \(y_0\) is, \(y(t)=-\infty\) for some finite value of \(t\).
4. Verify your results experiments analytically. Start by separating variables in (A) to obtain
  
  \[ {y'\over ay-by^2-q}=1. \]
  
  To decide what to do next you’ll have to use the quadratic formula. This should lead you to see why there are three cases. Take it from there!
  
  Because of its role in the transition between these three cases, \(q_0=a^2/4b\) is called a bifurcation value of \(q\). In general, if \(q\) is a parameter in any differential equation, \(q_0\) is said to be a bifurcation value of \(q\) if the nature of the solutions of the equation with \(q<q_0\) is qualitatively different from the nature of the solutions with \(q>q_0\).
L By plotting direction fields and solutions of

\[ y'=qy-y^3, \]

convince yourself that \(q_0=0\) is a bifurcation value of \(q\) for this equation. Explain what makes you draw this conclusion.
Suppose a disease spreads according to the model of Exercise 29, but there’s a medication that cures the infected population at a constant rate of \(q\) individuals per unit time, where \(q>0\). Then the equation for the number of infected individuals becomes

\[ I'=rI(S-I)-q. \]

Assuming that \(I(0)=I_0>0\), use the results of Exercise 31 to describe what happens as \(t\to\infty\).
Assuming that \(p \not\equiv 0\), state conditions under which the linear equation

\[ y'+p(x)y=f(x) \]

is separable. If the equation satisfies these conditions, solve it by separation of variables and by the method developed in Section 2.1.

Show answer

\(f=ap, \quad \mbox{where \)a\(=constant}\)

Solve the equations in Exercises 35–38 using variation of parameters followed by separation of variables.

\(\dst{y'+y={2xe^{-x}\over1+ye^x}}\)

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\(y=e^{-x}\left(-1\pm\sqrt{2x^2+c}\right)\)
\(\dst{xy'-2y={x^6\over y+x^2}}\)

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\(y=x^2\left(-1+\sqrt{x^2+c}\right)\)
\(\dst{y'-y}={(x+1)e^{4x}\over(y+e^x)^2}\)

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\(y=e^x\left(-1+(3xe^x+c)^{1/3}\right)\)
\(y'-2y=\dst{xe^{2x}\over1-ye^{-2x}}\)

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\(y=e^{2x}(1\pm\sqrt{c-x^2})\)
Use variation of parameters to show that the solutions of the following equations are of the form \(y=uy_1\), where \(u\) satisfies a separable equation \(u'=g(x)p(u)\). Find \(y_1\) and \(g\) for each equation.

(a) \(xy'+y=h(x)p(xy)\) (b) \(\dst{xy'-y=h(x) p\left({y\over x}\right)}\)

(c) \(y'+y=h(x) p(e^xy)\) (d) \(xy'+ry=h(x) p(x^ry)\)

(e) \(\dst{y'+{v'(x)\over v(x)}y= h(x) p\left(v(x)y\right)}\)

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(a) \(y_1=1/x\); \(g(x)=h(x)\)

(b) \(y_1=x\); \(g(x)=h(x)/x^2\) (c) \(y_1=e^{-x}\); \(g(x)=e^xh(x)\)

(d) \(y_1=x^{-r}\); \(g(x)=x^{r-1}h(x)\) (e) \(y_1=1/v(x)\); \(g(x)=v(x)h(x)\)

(a) \(xy'+y=h(x)p(xy)\)	(b) \(\dst{xy'-y=h(x) p\left({y\over x}\right)}\)
(c) \(y'+y=h(x) p(e^xy)\)	(d) \(xy'+ry=h(x) p(x^ry)\)
(e) \(\dst{y'+{v'(x)\over v(x)}y= h(x) p\left(v(x)y\right)}\)