2.1 Linear First Order Equations

A first order differential equation is said to be linear if it can be written as

A first order differential equation that can’t be written like this is nonlinear. We say that (2.1.1) is homogeneous if \(f\equiv0\); otherwise it’s nonhomogeneous. Since \(y\equiv0\) is obviously a solution of the homgeneous equation

we call it the trivial solution. Any other solution is nontrivial.

General Solution of a Linear First Order Equation

To motivate a definition that we’ll need, consider the simple linear first order equation

From calculus we know that \(y\) satisfies this equation if and only if

where \(c\) is an arbitrary constant. We call \(c\) a parameter and say that (2.1.3) defines a one–parameter family of functions. For each real number \(c\), the function defined by (2.1.3) is a solution of (2.1.2) on \((-\infty,0)\) and \((0,\infty)\); moreover, every solution of (2.1.2) on either of these intervals is of the form (2.1.3) for some choice of \(c\). We say that (2.1.3) is the general solution of (2.1.2).

We’ll see that a similar situation occurs in connection with any first order linear equation

that is, if \(p\) and \(f\) are continuous on some open interval \((a,b)\) then there’s a unique formula \(y=y(x,c)\) analogous to (2.1.3) that involves \(x\) and a parameter \(c\) and has the these properties:

• For each fixed value of \(c\), the resulting function of \(x\) is a solution of (2.1.4) on \((a,b)\).

• If \(y\) is a solution of (2.1.4) on \((a,b)\), then \(y\) can be obtained from the formula by choosing \(c\) appropriately.

We’ll call \(y=y(x,c)\) the general solution of (2.1.4).

When this has been established, it will follow that an equation of the form

has a general solution on any open interval \((a,b)\) on which \(P_0\), \(P_1\), and \(F\) are all continuous and \(P_0\) has no zeros, since in this case we can rewrite (2.1.5) in the form (2.1.4) with \(p=P_1/P_0\) and \(f=F/P_0\), which are both continuous on \((a,b)\).

To avoid awkward wording in examples and exercises, we won’t specify the interval \((a,b)\) when we ask for the general solution of a specific linear first order equation. Let’s agree that this always means that we want the general solution on every open interval on which \(p\) and \(f\) are continuous if the equation is of the form (2.1.4), or on which \(P_0\), \(P_1\), and \(F\) are continuous and \(P_0\) has no zeros, if the equation is of the form (2.1.5). We leave it to you to identify these intervals in specific examples and exercises.

For completeness, we point out that if \(P_0\), \(P_1\), and \(F\) are all continuous on an open interval \((a,b)\), but \(P_0\) does have a zero in \((a,b)\), then (2.1.5) may fail to have a general solution on \((a,b)\) in the sense just defined. Since this isn’t a major point that needs to be developed in depth, we won’t discuss it further; however, see Exercise 44 for an example.

Homogeneous Linear First Order Equations

We begin with the problem of finding the general solution of a homogeneous linear first order equation. The next example recalls a familiar result from calculus.

Solution (a) You already know from calculus that if \(c\) is any constant, then \(y=ce^{ax}\) satisfies (2.1.6). However, let’s pretend you’ve forgotten this, and use this problem to illustrate a general method for solving a homogeneous linear first order equation.

We know that (2.1.6) has the trivial solution \(y\equiv0\). Now suppose \(y\) is a nontrivial solution of (2.1.6). Then, since a differentiable function must be continuous, there must be some open interval \(I\) on which \(y\) has no zeros. We rewrite (2.1.6) as

for \(x\) in \(I\). Integrating this shows that

where \(k\) is an arbitrary constant. Since \(e^{ax}\) can never equal zero, \(y\) has no zeros, so \(y\) is either always positive or always negative. Therefore we can rewrite \(y\) as

where

This shows that every nontrivial solution of (2.1.6) is of the form \(y=ce^{ax}\) for some nonzero constant \(c\). Since setting \(c=0\) yields the trivial solution, all solutions of (2.1.6) are of the form (2.1.7). Conversely, (2.1.7) is a solution of (2.1.6) for every choice of \(c\), since differentiating (2.1.7) yields \(y'=ace^{ax}=ay\).

Solution (b) Imposing the initial condition \(y(x_0)=y_0\) yields \(y_0=ce^{ax_0}\), so \(c=y_0e^{-ax_0}\) and

Figure 2.1.1 show the graphs of this function with \(x_{0}=0\), \(y_{0}=1\), and various values of \(a\).

Solution (a) We rewrite (2.1.8) as

where \(x\) is restricted to either \((-\infty,0)\) or \((0,\infty)\). If \(y\) is a nontrivial solution of (2.1.10), there must be some open interval I on which \(y\) has no zeros. We can rewrite (2.1.10) as

for \(x\) in \(I\). Integrating shows that

Since a function that satisfies the last equation can’t change sign on either \((-\infty,0)\) or \((0,\infty)\), we can rewrite this result more simply as

where

We’ve now shown that every solution of (2.1.10) is given by (2.1.11) for some choice of \(c\). (Even though we assumed that \(y\) was nontrivial to derive (2.1.11), we can get the trivial solution by setting \(c=0\) in (2.1.11).) Conversely, any function of the form (2.1.11) is a solution of (2.1.10), since differentiating (2.1.11) yields

and substituting this and (2.1.11) into (2.1.10) yields

Figure 2.1.2 shows the graphs of some solutions corresponding to various values of \(c\)

Solution (b) Imposing the initial condition \(y(1)=3\) in (2.1.11) yields \(c=3\). Therefore the solution of (2.1.9) is

The interval of validity of this solution is \((0,\infty)\).

The results in Examples 2.1.3(a) and 2.1.4(b) are special cases of the next theorem.

Proof If \(y=ce^{-P(x)}\), differentiating \(y\) and using (2.1.14) shows that

so \(y'+p(x)y=0\); that is, \(y\) is a solution of (2.1.12), for any choice of \(c\).

Now we’ll show that any solution of (2.1.12) can be written as \(y=ce^{-P(x)}\) for some constant \(c\). The trivial solution can be written this way, with \(c=0\). Now suppose \(y\) is a nontrivial solution. Then there’s an open subinterval \(I\) of \((a,b)\) on which \(y\) has no zeros. We can rewrite (2.1.12) as

for \(x\) in \(I\). Integrating (2.1.15) and recalling (2.1.13) yields

where \(k\) is a constant. This implies that

Since \(P\) is defined for all \(x\) in \((a,b)\) and an exponential can never equal zero, we can take \(I=(a,b)\), so \(y\) has zeros on \((a,b)\) \((a,b)\), so we can rewrite the last equation as \(y=ce^{-P(x)}\), where

Linear Nonhomogeneous First Order Equations

We’ll now solve the nonhomogeneous equation

When considering this equation we call

the complementary equation.

We’ll find solutions of (2.1.16) in the form \(y=uy_1\), where \(y_1\) is a nontrivial solution of the complementary equation and \(u\) is to be determined. This method of using a solution of the complementary equation to obtain solutions of a nonhomogeneous equation is a special case of a method called variation of parameters, which you’ll encounter several times in this book. (Obviously, \(u\) can’t be constant, since if it were, the left side of (2.1.16) would be zero. Recognizing this, the early users of this method viewed \(u\) as a “parameter” that varies; hence, the name “variation of parameters.”)

If

Substituting these expressions for \(y\) and \(y'\) into (2.1.16) yields

which reduces to

since \(y_1\) is a solution of the complementary equation; that is,

In the proof of Theorem 2.2.1 we saw that \(y_1\) has no zeros on an interval where \(p\) is continuous. Therefore we can divide (2.1.17) through by \(y_1\) to obtain

We can integrate this (introducing a constant of integration), and multiply the result by \(y_1\) to get the general solution of (2.1.16). Before turning to the formal proof of this claim, let’s consider some examples.

By applying (a) of Example  2.1.3 with \(a=-2\), we see that \(y_1=e^{-2x}\) is a solution of the complementary equation \(y'+2y=0\). Therefore we seek solutions of (2.1.18) in the form \(y=ue^{-2x}\), so that

Therefore \(y\) is a solution of (2.1.18) if and only if

Therefore

and

is the general solution of (2.1.18).

Figure 2.1.3 shows a direction field and some integral curves for (2.1.18).

Solution (a) Here \(p(x)=\cot x\) and \(f(x)= x\csc x\) are both continuous except at the points \(x=r\pi\), where \(r\) is an integer. Therefore we seek solutions of (2.1.20) on the intervals \(\left(r\pi, (r+1)\pi \right)\). We need a nontrival solution \(y_1\) of the complementary equation; thus, \(y_1\) must satisfy \(y_1'+(\cot x)y_1=0\), which we rewrite as

Integrating this yields

where we take the constant of integration to be zero since we need only one function that satisfies (2.1.22). Clearly \(y_1=1/\sin x\) is a suitable choice. Therefore we seek solutions of (2.1.20) in the form

so that

and

Therefore \(y\) is a solution of (2.1.20) if and only if

Integrating this yields

is the general solution of (2.1.20) on every interval \(\left(r\pi,(r+1)\pi\right)\) (\(r=\)integer).

Solution (b) Imposing the initial condition \(y(\pi/2)=1\) in (2.1.25) yields

Thus,

is a solution of (2.1.21). The interval of validity of this solution is \((0,\pi)\); Figure 2.1.4 shows its graph.

We summarize the method of variation of parameters for solving

as follows:

1. Find a function \(y_1\) such that

   \[ {y_1'\over y_1}=-p(x). \]

   For convenience, take the constant of integration to be zero.

2. Write

   \begin{equation} y=uy_1 \tag{2.1.27}\end{equation}

   to remind yourself of what you’re doing.

3. Write \(u'y_1=f\) and solve for \(u'\); thus, \(u'=f/y_1\).

4. Integrate \(u'\) to obtain \(u\), with an arbitrary constant of integration.

5. Substitute \(u\) into (2.1.27) to obtain \(y\).

To solve an equation written as

we recommend that you divide through by \(P_0(x)\) to obtain an equation of the form (2.1.26) and then follow this procedure.

Solutions in Integral Form

Sometimes the integrals that arise in solving a linear first order equation can’t be evaluated in terms of elementary functions. In this case the solution must be left in terms of an integral.

Solution (a) To apply variation of parameters, we need a nontrivial solution \(y_1\) of the complementary equation; thus, \(y_1'-2xy_1=0\), which we rewrite as

Integrating this and taking the constant of integration to be zero yields

We choose \(y_1=e^{x^2}\) and seek solutions of (2.1.28) in the form \(y=ue^{x^2}\), where

Therefore

but we can’t simplify the integral on the right because there’s no elementary function with derivative equal to \(e^{-x^2}\). Therefore the best available form for the general solution of (2.1.28) is

Solution (b) Since the initial condition in (2.1.28) is imposed at \(x_0=0\), it is convenient to rewrite (2.1.29) as

Setting \(x=0\) and \(y=y_0\) here shows that \(c=y_0\). Therefore the solution of the initial value problem is

For a given value of \(y_0\) and each fixed \(x\), the integral on the right can be evaluated by numerical methods. An alternate procedure is to apply the numerical integration procedures discussed in Chapter 3 directly to the initial value problem (2.1.28). Figure 2.1.5 shows graphs of of (2.1.30) for several values of \(y_0\).

An Existence and Uniqueness Theorem

The method of variation of parameters leads to this theorem.

Proof (a) To show that (2.1.32) is the general solution of (2.1.31) on \((a,b)\), we must prove that:

1. If \(c\) is any constant, the function \(y\) in (2.1.32) is a solution of (2.1.31) on \((a,b)\).

2. If \(y\) is a solution of (2.1.31) on \((a,b)\) then \(y\) is of the form (2.1.32) for some constant \(c\).

To prove (i), we first observe that any function of the form (2.1.32) is defined on \((a,b)\), since \(p\) and \(f\) are continuous on \((a,b)\). Differentiating (2.1.32) yields

Since \(y_1'=-p(x)y_1\), this and (2.1.32) imply that

which implies that \(y\) is a solution of (2.1.31).

To prove (ii), suppose \(y\) is a solution of (2.1.31) on \((a,b)\). From the proof of Theorem 2.1.1, we know that \(y_1\) has no zeros on \((a,b)\), so the function \(u=y/y_1\) is defined on \((a,b)\). Moreover, since

Integrating \(u'=f/y_1\) yields

which implies (2.1.32), since \(y=uy_1\).

(b) We’ve proved (a), where \(\int f(x)/y_1(x)\,dx\) in (2.1.32) is an arbitrary antiderivative of \(f/y_1\). Now it’s convenient to choose the antiderivative that equals zero when \(x=x_0\), and write the general solution of (2.1.31) as

Since

we see that \(y(x_0)=y_0\) if and only if \(c=y_0/y_1(x_0)\).

2.1 Exercises

In Exercises 1–5 find the general solution.

1. \(y'+ay=0\) (\(a\)=constant)

   Show answer

   \(y=e^{-ax}\)

2. \(y'+3x^2y=0\)

   Show answer

   \(y=ce^{-x^3}\)

3. \(xy'+(\ln x)y=0\)

   Show answer

   \(y=ce^{-(\ln x)^2/2}\)

4. \(xy'+3y=0\)

   Show answer

   \(\dst{y={c\over x^3}}\)

5. \(x^2y'+y=0\)

   Show answer

   \(y=ce^{1/x}\)

6. \(\dst{y'+\left({1+x\over x}\right)y=0,\quad y(1)=1}\)

   Show answer

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

7. \(\dst{xy'+\left(1+{1\over\ln x}\right)y=0,\quad y(e)=1}\)

   Show answer

   \(\dst{y={e\over x\ln x}}\)

8. \(\dst{xy'+(1+ x\cot x)y=0,\quad y\left({\pi\over 2} \right)=2}\)

   Show answer

   \(\dst{y={\pi\over x\sin x}}\)

9. \(\dst{y'-\left({2x\over 1+x^2}\right)y=0,\quad y(0)=2}\)

   Show answer

   \(y=2(1+x^2)\)

10. \(\dst{y'+{k\over x}y=0,\quad y(1)=3 \quad\mbox{(\)k\(= constant)}}\)

    Show answer

    \(y=3x^{-k}\)

11. \(\dst y'+(\tan kx)y=0,\quad y(0)=2 \quad\mbox{(\)k=\( constant)}\)

    Show answer

    \(y=c(\cos kx)^{1/k}\)

12. C/G \(y'+3y=1\)

    Show answer

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

13. C/G \(\dst{y'+\left({1\over x}- 1\right)y=-{2\over x}}\)

    Show answer

    \(\dst{y={2\over x}+{c\over x}e^x}\)

14. C/G \(y'+2xy=xe^{-x^2}\)

    Show answer

    \(\dst{y=e^{-x^2}\left({x^2\over2}+c\right)}\)

15. C/G \(\dst{y'+{2x\over1+x^2}y={e^{-x}\over1+x^2}}\)

    Show answer

    \(y=-\dst{e^{-x}+c\over1+x^2}\)

16. \(\dst{y'+{1\over x}y={7\over x^2}+3}\)

    Show answer

    \(\dst{y={7\ln |x|\over x}+{3\over 2}x + {c\over x}}\)

17. \(\dst{y'+{4\over x-1}y = {1\over (x-1)^5}+{\sin x\over (x-1)^4}}\)

    Show answer

    \(y=(x-1)^{-4}(\ln |x-1| -\cos x+c)\)

18. \(xy'+(1+2x^2)y=x^3e^{-x^2}\)

    Show answer

    \(\dst{y=e^{-x^2}\left({x^3 \over 4} + {c\over x}\right)}\)

19. \(\dst{xy'+2y={2\over x^2}+1}\)

    Show answer

    \(\dst{y={2\ln |x|\over x^2} + {1\over 2} + {c\over x^2}}\)

20. \(y'+(\tan x)y=\cos x\)

    Show answer

    \(y=(x+c)\cos x\)

21. \(\dst{(1+x)y'+2y={\sin x \over 1 + x}}\)

    Show answer

    \(\dst{y={c-\cos x\over (1+x)^2}}\)

22. \((x-2)(x-1)y'-(4x-3)y=(x-2)^3\)

    Show answer

    \(\dst{y=- {1\over 2}{(x-2)^3\over (x-1)}+c{(x-2)^5\over (x-1)}}\)

23. \(y'+(2\sin x\cos x) y=e^{-\sin^2x}\)

    Show answer

    \(y=(x+c)e^{-\sin^2x}\)

24. \(x^2y'+3xy=e^x\)

    Show answer

    \(y=\dst{\frac{e^{x}}{x^{2}}-\frac{e^{x}}{x^{3}}+\frac{c}{x^{2}}}\).

25. C/G \(y'+7y=e^{3x},\quad y(0)=0\)

    Show answer

    \(y=\dst{\frac{e^{3x}-e^{-7x}}{10}}\)

26. C/G \(\dst{(1+x^2)y'+4xy={2\over 1+x^2},\quad y(0)=1}\)

    Show answer

    \(\dst{\frac{2x+1}{(1+x^{2})^{2}}}\)

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

    Show answer

    \(y=\dst{\frac{1}{x^{2}}\ln\left(\frac{1+x^{2}}{2}\right)}\)

28. C/G \(\dst{y'+ (\cot x)y=\cos x,\quad y\left({\pi\over 2}\right)=1}\)

    Show answer

    \(\dst{y={1\over 2} (\sin x +\csc x)}\)

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

    Show answer

    \(\dst{y=\frac{2\ln |x|}{x} + \frac{x}{2} - \frac{1}{2x}}\)

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

    Show answer

    \(y=(x-1)^{-3} \left[\ln (1-x)-\cos x\right]\)

31. \(xy'+2y=8x^2,\quad y(1)=3\)

    Show answer

    \(\dst{y=2x^2 + {1\over x^2} \quad (0,\infty)}\)

32. \(xy'-2y=-x^2,\quad y(1)=1\)

    Show answer

    \(y=x^2(1-\ln x)\)

33. \(y'+2xy=x,\quad y(0)=3\)

    Show answer

    \(\dst{y={1 \over 2}+{5\over 2}e^{-x^2}}\)

34. \(\dst{(x-1)y'+3y={1+(x-1)\sec^2x\over (x-1)^3},\quad y(0)=-1}\)

    Show answer

    \(\dst{y={\ln|x-1|+\tan x+1\over(x-1)^3}}\)

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

    Show answer

    \(\dst{y={\ln|x|+x^2+1\over(x+2)^4}}\)

36. \((x^2-1)y'-2xy=x(x^2-1),\quad y(0)=4\)

    Show answer

    \(\dst{y=(x^2-1)\left(\displaystyle{{1\over2} \ln |x^2-1|-4}\right)}\)

37. \((x^2-5)y'-2xy=-2x(x^2-5),\quad y(2)=7\)

    Show answer

    \(\dst{y=-(x^2-5)\left(7+\ln|x^2-5|\right)}\)

38. \(y'+2xy=x^2,\quad y(0)=3\)

    Show answer

    \(\dst{y=e^{-x^2}\left(3 +\int^x_0 t^2e^{t^2}\, dt\right)}\)

39. \(\dst{y'+{1\over x}y={\sin x\over x^2},\quad y(1)=2}\)

    Show answer

    \(\dst{y={1\over x}\left(2 +\int^x_1{\sin t\over t}\, dt\right)}\)

40. \(\dst{y'+y={e^{-x}\tan x\over x},\quad y(1)=0}\)

    Show answer

    \(\dst{y=e^{-x} \int^x_1 {\tan t\over t}\, dt}\)

41. \(\dst{y'+{2x\over 1+x^2}y={e^x\over (1+x^2)^2}, \quad y(0)=1}\)

    Show answer

    \(\dst{y={1\over 1+x^2}\left(1+\int^x_0 {e^t\over 1+t^2}\, dt\right)}\)

42. \(xy'+(x+1)y=e^{x^2},\quad y(1)=2\)

    Show answer

    \(\dst{y={1\over x}\left(2e^{-(x-1)}+e^{-x} \int_1^x e^te^{t^2}\,dt\right)}\)

43. Experiments indicate that glucose is absorbed by the body at a rate proportional to the amount of glucose present in the bloodstream. Let \(\lambda\) denote the (positive) constant of proportionality. Now suppose glucose is injected into a patient’s bloodstream at a constant rate of \(r\) units per unit of time. Let \(G=G(t)\) be the number of units in the patient’s bloodstream at time \(t>0\). Then

    \[ G'=-\lambda G+r, \]

    where the first term on the right is due to the absorption of the glucose by the patient’s body and the second term is due to the injection. Determine \(G\) for \(t>0\), given that \(G(0)=G_0\). Also, find \(\lim_{t\to\infty}G(t)\).

    Show answer

    \(G=\dst{{r\over\lambda}+\left(G_0-{r\over\lambda}\right)e^{-\lambda t}}\)  \(\lim_{t\to\infty}G(t)=\dst{r\over\lambda}\)

44. 1. L Plot a direction field and some integral curves for

       \[ xy'-2y=-1 \tag*{\rm (A)} \]

       on the rectangular region \(\{-1\le x\le 1, -.5\le y\le 1.5\}\). What do all the integral curves have in common?

    2. Show that the general solution of (A) on \((-\infty,0)\) and \((0,\infty)\) is

       \[ y={1\over2}+cx^2. \]

    3. Show that \(y\) is a solution of (A) on \((-\infty,\infty)\) if and only if

       \[ y=\left\{\begin{array}{ll}\dst{{1\over2}+c_1x^2}, &x \ge 0,\\[6pt] \dst{{1\over2}+c_2x^2}, &x < 0,\end{array}\right. \]

       where \(c_1\) and \(c_2\) are arbitrary constants.

    4. Conclude from (c) that all solutions of (A) on \((-\infty,\infty)\) are solutions of the initial value problem

       \[ xy'-2y=-1,\quad y(0)={1\over2}. \]

    5. Use (b) to show that if \(x_0\ne0\) and \(y_0\) is arbitrary, then the initial value problem

       \[ xy'-2y=-1,\quad y(x_0)=y_0 \]

       has infinitely many solutions on (\(-\infty,\infty\)). Explain why this does’nt contradict Theorem 2.1.1(b).

45. Suppose \(f\) is continuous on an open interval \((a,b)\) and \(\alpha\) is a constant.

    1. Derive a formula for the solution of the initial value problem

       \[ y'+\alpha y=f(x),\quad y(x_0)=y_0, \tag*{\rm (A)} \]

       where \(x_0\) is in \((a,b)\) and \(y_0\) is an arbitrary real number.

    2. Suppose \((a,b)=(a,\infty)\), \(\alpha > 0\) and \(\displaystyle{\lim_{x\to\infty} f(x)=L}\). Show that if \(y\) is the solution of (A), then \(\displaystyle{\lim_{x\to \infty} y(x)=L/\alpha}\).

    Show answer

    (a) \(\dst{y=y_0 e^{-a(x-x_0)}+ e^{-ax} \int^x_{x_0} e^{at}f(t)\, dt}\)

46. Assume that all functions in this exercise are defined on a common interval \((a,b)\).

    1. Prove: If \(y_1\) and \(y_2\) are solutions of

       \[ y'+p(x)y=f_1(x) \]

       and

       \[ y'+p(x)y=f_2(x) \]

       respectively, and \(c_1\) and \(c_2\) are constants, then \(y=c_1y_1+c_2y_2\) is a solution of

       \[ y'+p(x)y=c_1f_1(x)+c_2f_2(x). \]

       (This is the principle of superposition.)

    2. Use (a) to show that if \(y_1\) and \(y_2\) are solutions of the nonhomogeneous equation

       \[ y'+p(x)y=f(x), \tag*{\rm (A)} \]

       then \(y_1-y_2\) is a solution of the homogeneous equation

       \[ y'+p(x)y=0. \tag*{\rm (B)} \]

    3. Use (a) to show that if \(y_1\) is a solution of (A) and \(y_2\) is a solution of (B), then \(y_1+y_2\) is a solution of (A).

47. Some nonlinear equations can be transformed into linear equations by changing the dependent variable. Show that if

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

    where \(y\) is a function of \(x\) and \(g\) is a function of \(y\), then the new dependent variable \(z=g(y)\) satisfies the linear equation

    \[ z'+p(x)z=f(x). \]

48. Solve by the method discussed in Exercise 47.

    (a) \((\sec^2y)y'- 3\tan y=-1\)

    (b) \(\dst{e^{y^2}\left(2yy'+ {2\over x}\right) ={1\over x^2}}\)

    (c) \(\dst{{xy'\over y} + 2\ln y=4x^2}\)

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

    Show answer

    (a) \(\dst{y=\tan^{-1}\left({1\over 3} + ce^{3x}\right)}\)   (b) \(y=\pm\dst{\left[\ln\left(\dst{1\over x}+\dst{c\over x^2}\right)\right]^{1/2}}\)

        (c) \(\dst{y=\exp\left(x^2 + {c\over x^2}\right)}\)   (d) \(\dst{y=-1 + {x\over c + 3\ln |x|}}\)

49. We’ve shown that if \(p\) and \(f\) are continuous on \((a,b)\) then every solution of

    \[ y'+p(x)y=f(x) \tag*{\rm(A)} \]

    on \((a,b)\) can be written as \(y=uy_1\), where \(y_1\) is a nontrivial solution of the complementary equation for (A) and \(u'=f/y_1\). Now suppose \(f\), \(f'\), …, \(f^{(m)}\) and \(p\), \(p'\), …, \(p^{(m-1)}\) are continuous on \((a,b)\), where \(m\) is a positive integer, and define

    \begin{eqnarray*} f_0&=&f, \\ f_j&=&f_{j-1}'+pf_{j-1},\quad 1\le j\le m. \end{eqnarray*}

    Show that

    \[ u^{(j+1)}={f_j\over y_1},\quad 0\le j\le m. \]