DIFFERENTIAL EQUATIONS

Chapter 9

Differential Equations

Big idea: many laws of nature do not tell you a quantity directly. They tell you how fast it changes. A differential equation states such a law, and solving it recovers the quantity itself.

A cup of hot drink stands in a room at $20°$C. You notice that it cools quickly at first, and then more and more slowly. The hotter the drink is compared with the room, the faster it loses heat. A simple law says this exactly: the rate of cooling is proportional to the temperature difference. In symbols, with $T$ the temperature after $t$ minutes,

$$\frac{dT}{dt} = -0.1\,(T - 20).$$

This is a differential equation: an equation that contains the derivative of an unknown function. It does not give $T$. It gives a rule about how $T$ changes. To find $T$, you have to solve the equation. If the drink starts at $80°$C, the solution is

$$T(t) = 20 + 60\,e^{-0.1t}.$$

You can check this by differentiating: $\frac{dT}{dt} = -6\,e^{-0.1t}$, and $-0.1\,(T - 20) = -0.1 \cdot 60\,e^{-0.1t} = -6\,e^{-0.1t}$. They match. After $10$ minutes the temperature is $T(10) = 20 + 60e^{-1} \approx 42.07°$C. The computer confirmed this by solving the equation step by step on its own, without using the formula.

You already know the simplest kind. In Chapter 7 you solved $\frac{dy}{dx} = 2x$ by integrating: $y = x^2 + C$. Every integral is the solution of a differential equation of the form $\frac{dy}{dx} = f(x)$. The equations in this chapter are harder, because the right side may contain $y$ too, as in $\frac{dy}{dx} = x\,y$. You will need all your integration skills.

Differential equations appear wherever something changes according to a rule: cooling and heating, the growth of a population, a decaying radioactive sample, the motion of a falling object, the charging of a battery, the spread of a rumour.

A differential equation is an equation that contains an unknown function of one variable together with its derivatives. We write the derivatives as $\frac{dy}{dx}$, $\frac{d^2y}{dx^2}$, ... or in the short form $y'$, $y''$, ... Here are some examples:

• $\dfrac{dy}{dx} = 2y$ (the rate of growth is twice the value),
• $y'' + y = 0$ (the second derivative is the negative of the function),
• $x\,y' = y + x$.

In this chapter there is only one independent variable (such as $x$ or $t$), so these are ordinary differential equations.

A solution of a differential equation is a function that makes the equation true when you put it in. To test a candidate, calculate its derivatives and substitute.

Test 1. Is $y = 3e^{2x}$ a solution of $y' = 2y$? Differentiate: $y' = 6e^{2x}$. And $2y = 6e^{2x}$. They are equal for every $x$, so yes. (At $x = 0$ both sides equal $6$.) What about $y = e^{3x}$? Then $y' = 3e^{3x}$ but $2y = 2e^{3x}$, which is different, so no. The computer checked both claims at several values of $x$.

Test 2. Is $y = 2\cos x + 5\sin x$ a solution of $y'' + y = 0$? We get $y' = -2\sin x + 5\cos x$ and $y'' = -2\cos x - 5\sin x$, which is exactly $-y$. So $y'' + y = 0$. Yes.

Take $y' = 2y$ again. The function $y = C\,e^{2x}$ is a solution for every number $C$. Each value of $C$ gives one curve, and every curve in the picture satisfies the same equation. The equation does not choose between them. To choose one, you need extra information, such as the value of $y$ at one point. That is the topic of Section 9.3.

The order of a differential equation is the order of the highest derivative that appears in it. First derivative: order $1$. Second derivative: order $2$. And so on. Only the highest derivative counts. The unknown function itself (order $0$) and any lower derivatives do not change the order.

The degree is defined only when the equation is a polynomial in its derivatives. Then the degree is the power of the highest-order derivative.

Before you read off the degree, clear any roots or fractions that contain derivatives, so that the equation is a polynomial in $y', y'', \dots$. Then look at the highest derivative and find its power.

Notice two things. First, in $\sin(y') + y = x$ the derivative sits inside a sine function. A sine is not a polynomial, so the degree is not defined. Second, the power of $y$ itself does not matter: $y' + y^5 = 0$ has degree $1$ because $y'$ appears to the first power.

State the order and degree of each equation.

1. $y''' + (y')^4 + y = 0$
2. $y' + e^{y'} = x$
3. $y'' + (y')^3 + 6y = 0$

Answers

1. Order 3, degree 1. The highest derivative is $y'''$, to the first power.
2. Order 1, degree not defined. The derivative is inside an exponential, which is not a polynomial.
3. Order 2, degree 1. The highest derivative is $y''$ (power $1$). The cube on $y'$ is a lower derivative, so it does not matter.

1. State the order and the degree (if defined) of $\dfrac{dy}{dx} + y = e^x$.
2. State the order and the degree (if defined) of $y'' + (y')^3 + 6y = 0$.
3. State the order and the degree (if defined) of $(y''')^2 + (y'')^3 + (y')^4 + y^5 = 0$.
4. State the order and the degree (if defined) of $y''' + 2y'' + y' = 0$.
5. State the order and the degree (if defined) of $(y')^2 + x\,y = 0$.
6. State the order and the degree (if defined) of $y'' = \cos 3x + \sin 3x$.
7. State the order and the degree (if defined) of $\sin(y'') + y = 0$.
8. State the order and the degree (if defined) of $x\,(y'')^2 + y\,(y')^3 + x^3 = 0$.
9. State the order and the degree (if defined) of $y' = \sqrt{1 + (y'')^2}$. (Square both sides first.)
10. State the order and the degree (if defined) of $y'' + e^{y'} = 0$.

Answers

1. Order 1, degree 1.
2. Order 2, degree 1.
3. Order 3, degree 2.
4. Order 3, degree 1.
5. Order 1, degree 2.
6. Order 2, degree 1.
7. Order 2, degree not defined.
8. Order 2, degree 2.
9. Order 2, degree 2.
10. Order 2, degree 1.

Look at the simplest equation $\dfrac{dy}{dx} = 2x$. Integrating gives $y = x^2 + C$. Here $C$ is an arbitrary constant: you may choose any number. The result is the general solution. It describes all the solutions in one formula.

A particular solution is one single solution, obtained by giving the arbitrary constants definite values. Usually you do this using a condition, for example "the curve passes through the point $(1, 3)$". Then $3 = 1^2 + C$, so $C = 2$, and the particular solution is $y = x^2 + 2$.

The general solution of an equation of order $n$ has $n$ independent arbitrary constants. The order tells you how many.

• Order $1$: one constant. Example: $y' = 2y$ has general solution $y = C\,e^{2x}$.
• Order $2$: two constants. Example: $y'' + y = 0$ has general solution $y = A\cos x + B\sin x$. The computer checked, for several choices of $A$ and $B$, that this is a solution.

A second-order equation needs two conditions. For $y'' + y = 0$, suppose we are told $y(0) = 1$ and $y'(0) = 2$. Since $y = A\cos x + B\sin x$ and $y' = -A\sin x + B\cos x$, putting $x = 0$ gives $A = 1$ and $B = 2$. So the particular solution is $y = \cos x + 2\sin x$. Check: $y(0) = 1$ and $y'(0) = 2$.

A ball is thrown straight up. Its height $h$ (in metres) after $t$ seconds follows $h'' = -10$ (gravity slows it down by $10$ m/s every second). Integrate once: $h' = -10t + C_1$. Integrate again: $h = -5t^2 + C_1 t + C_2$. If the ball starts at ground level with speed $20$ m/s, then $h(0) = 0$ gives $C_2 = 0$, and $h'(0) = 20$ gives $C_1 = 20$. So $h = 20t - 5t^2$. The ball is highest when $h' = 20 - 10t = 0$, that is at $t = 2$, where $h = 20$ m. It is back on the ground at $t = 4$ s.

Start from the family $y = C\,e^{2x}$. Differentiate: $y' = 2C\,e^{2x}$. The right side is $2y$, so $y' = 2y$. The constant has disappeared, and we have found the differential equation that this family satisfies.

1. Show that $y = 2x + \dfrac{3}{x}$ is a solution of $x\,y' + y = 4x$ for $x > 0$.
2. Show that $y = C\,e^{-3x}$ is a solution of $y' + 3y = 0$. Find $C$ if $y(0) = 5$, and then find $y(1)$.
3. Show that $y = A\cos 2x + B\sin 2x$ is a solution of $y'' + 4y = 0$. Find the solution with $y(0) = 3$ and $y'(0) = -4$.
4. Show that $y = x\ln x$ is a solution of $y' = \dfrac{y}{x} + 1$ for $x > 0$.
5. Is $y = \sin x$ a solution of $y'' + y = 0$? Is $y = x^2$?
6. Show that $y = C_1 e^{x} + C_2 e^{2x}$ is a solution of $y'' - 3y' + 2y = 0$. Find the solution with $y(0) = 1$ and $y'(0) = 0$.
7. Find the particular solution of $y' = 3x^2 - 2x$ that satisfies $y(1) = 4$.
8. A population $P$ grows according to $\dfrac{dP}{dt} = 0.05\,P$, with $P(0) = 200$. Show that $P = 200\,e^{0.05t}$ and find $P(20)$.

Answers

1. $y' = 2 - \dfrac{3}{x^2}$, so $x\,y' = 2x - \dfrac{3}{x}$. Adding $y$: $2x - \dfrac{3}{x} + 2x + \dfrac{3}{x} = 4x$. It matches.
2. $y' = -3C\,e^{-3x} = -3y$, so $y' + 3y = 0$. From $y(0) = C = 5$. Then $y(1) = 5e^{-3} \approx 0.2489$.
3. $y'' = -4A\cos 2x - 4B\sin 2x = -4y$. Using $y(0) = A = 3$ and $y'(0) = 2B = -4$, we get $A = 3$ and $B = -2$. So $y = 3\cos 2x - 2\sin 2x$.
4. $y' = \ln x + 1$. And $\dfrac{y}{x} + 1 = \ln x + 1$. They are equal.
5. $y = \sin x$: $y'' = -\sin x$, so $y'' + y = 0$. Yes. $y = x^2$: $y'' = 2$, so $y'' + y = 2 + x^2 \ne 0$. No.
6. $y' = C_1 e^x + 2C_2 e^{2x}$ and $y'' = C_1 e^x + 4C_2 e^{2x}$. Then $y'' - 3y' + 2y = (1 - 3 + 2)C_1 e^x + (4 - 6 + 2)C_2 e^{2x} = 0$. From the conditions, $C_1 + C_2 = 1$ and $C_1 + 2C_2 = 0$, so $C_2 = -1$ and $C_1 = 2$. The solution is $y = 2e^x - e^{2x}$ (it gives $y(0) = 1$, $y'(0) = 0$).
7. Integrating, $y = x^3 - x^2 + C$. At $x = 1$: $0 + C = 4$, so $C = 4$. The solution is $y = x^3 - x^2 + 4$.
8. $P' = 0.05 \cdot 200\,e^{0.05t} = 0.05\,P$, and $P(0) = 200$. So it is the solution. $P(20) = 200e \approx 543.66$.

From here on we study equations of order $1$ and degree $1$. They can be written as $\dfrac{dy}{dx} = f(x, y)$. There is no single method that solves all of them, but three types cover most of what you will meet. Learn to recognise the type first.

An equation $\dfrac{dy}{dx} = f(x, y)$ is homogeneous if replacing $x$ by $\lambda x$ and $y$ by $\lambda y$ does not change $f$: $f(\lambda x, \lambda y) = f(x, y)$. This works exactly when $f$ can be written using only the ratio $\frac{y}{x}$. The computer tried several values of $\lambda$ on each function:

Some equations belong to more than one type. For instance $\frac{dy}{dx} = \frac{y}{x}$ is separable, homogeneous and linear. Use whichever method you find easiest.

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

$$\frac{dy}{dx} = f(x)\,g(y),$$

a product of a function of $x$ only and a function of $y$ only. Then you can move everything with $y$ to one side and everything with $x$ to the other, and integrate each side.

The steps:

1. Divide by $g(y)$ and "multiply by $dx$" to separate: $\dfrac{dy}{g(y)} = f(x)\,dx$.
2. Integrate both sides: $\displaystyle\int\frac{dy}{g(y)} = \int f(x)\,dx + C$. One constant is enough.
3. If you can, solve for $y$. If not, leave the answer as a relation between $x$ and $y$ (an implicit solution).

Solve $\dfrac{dy}{dx} = x\,y$. Separate: $\dfrac{dy}{y} = x\,dx$. Integrate: $\ln|y| = \dfrac{x^2}{2} + c$. Then $|y| = e^c\,e^{x^2/2}$, so

$$y = C\,e^{x^2/2}.$$

Here $C = \pm e^c$ may be positive or negative. The value $C = 0$ (the solution $y = 0$) also works. The computer checked this for $C = 2$ and $C = -3$.

Solve $\dfrac{dy}{dx} = \dfrac{1 + y^2}{1 + x^2}$. Separate: $\dfrac{dy}{1 + y^2} = \dfrac{dx}{1 + x^2}$. Integrate: $\tan^{-1}y = \tan^{-1}x + c$. Take the tangent of both sides and use the addition formula for tangent, writing $k = \tan c$:

$$y = \frac{x + k}{1 - kx}.$$

With $k = 2$, the computer confirmed that $y = \frac{x + 2}{1 - 2x}$ satisfies the equation at several points (away from $x = \frac{1}{2}$).

Solve $\dfrac{dy}{dx} = -\dfrac{x}{y}$. Separate: $y\,dy = -x\,dx$. Integrate: $\dfrac{y^2}{2} = -\dfrac{x^2}{2} + c$, which gives

$$x^2 + y^2 = C.$$

This is an implicit solution: a family of circles centred at the origin. The upper half of the circle with $C = 25$ is the function $y = \sqrt{25 - x^2}$, and the computer confirmed that it satisfies the equation, and that the slope of the circle at the points $(3, 4)$, $(4, 3)$, $(-3, 4)$ and $(0, 5)$ is $-\frac{x}{y}$.

Solve $\dfrac{dy}{dx} = y^2$. Separate: $\dfrac{dy}{y^2} = dx$. Integrate: $-\dfrac{1}{y} = x + c$, so $y = -\dfrac{1}{x + c}$. But we divided by $y^2$, which is not allowed if $y = 0$. And $y = 0$ is a solution of the equation (both sides are $0$). It cannot be obtained from $y = -\frac{1}{x + c}$ for any $c$. The computer confirmed that both $y = -\frac{1}{x + 2}$ and $y = 0$ are solutions.

Return to the drink of Section 9.1: $\dfrac{dT}{dt} = -0.1\,(T - 20)$, $T(0) = 80$. Separate: $\dfrac{dT}{T - 20} = -0.1\,dt$. Integrate: $\ln|T - 20| = -0.1t + c$. So $T = 20 + A\,e^{-0.1t}$. From $T(0) = 80$ we get $A = 60$. When does the drink reach $30°$C? Solve $60e^{-0.1t} = 10$: $t = 10\ln 6 \approx 17.92$ minutes. A step-by-step numerical solution gives $T = 30$ at that time.

1. Solve $\dfrac{dy}{dx} = \dfrac{y}{x}$ for $x > 0$.
2. Solve $\dfrac{dy}{dx} = e^{x - y}$.
3. Solve $(1 + x^2)\dfrac{dy}{dx} = x\,y$.
4. Solve $\dfrac{dy}{dx} = 2x\,(y + 1)$ with $y(0) = 1$.
5. Solve $\dfrac{dy}{dx} = y\cos x$ with $y(0) = 2$.
6. Solve $x\dfrac{dy}{dx} + y = 0$ for $x > 0$. What is the solution through $(1, 4)$?
7. Solve $\dfrac{dy}{dx} = \dfrac{y^2}{x}$ for $x > 0$. Is there a solution that this method misses?
8. A population of bacteria satisfies $\dfrac{dN}{dt} = 0.03\,N$, with $N(0) = 500$. Find $N(30)$.
9. Find the curve through $(0, 4)$ whose slope at every point is $-\dfrac{x}{y}$. (Take $y > 0$.)

Answers

1. Separate: $\dfrac{dy}{y} = \dfrac{dx}{x}$. So $\ln|y| = \ln x + c$, which gives $y = Cx$.
2. Write $e^{x - y} = e^x e^{-y}$. Separate: $e^y\,dy = e^x\,dx$. So $e^y = e^x + C$, and $y = \ln(e^x + C)$. (The computer checked this for $C = 1$.)
3. Separate: $\dfrac{dy}{y} = \dfrac{x}{1 + x^2}\,dx$. Then $\ln|y| = \dfrac{1}{2}\ln(1 + x^2) + c$, so $y = C\sqrt{1 + x^2}$.
4. Separate: $\dfrac{dy}{y + 1} = 2x\,dx$. So $\ln|y + 1| = x^2 + c$ and $y + 1 = Ce^{x^2}$. From $y(0) = 1$: $2 = C$. The solution is $y = 2e^{x^2} - 1$.
5. Separate: $\dfrac{dy}{y} = \cos x\,dx$. So $\ln|y| = \sin x + c$ and $y = Ce^{\sin x}$. From $y(0) = 2$: $C = 2$. The solution is $y = 2e^{\sin x}$.
6. Separate: $\dfrac{dy}{y} = -\dfrac{dx}{x}$. So $\ln|y| = -\ln x + c$, and $y = \dfrac{C}{x}$ (that is, $xy = C$). Through $(1, 4)$: $C = 4$, so $y = \dfrac{4}{x}$.
7. Separate: $\dfrac{dy}{y^2} = \dfrac{dx}{x}$. So $-\dfrac{1}{y} = \ln x + c$, which gives $y = -\dfrac{1}{\ln x + C}$. The method divides by $y^2$, and it misses the solution $y = 0$.
8. Separate: $\dfrac{dN}{N} = 0.03\,dt$, so $N = Ce^{0.03t}$, and $C = 500$. Then $N(30) = 500e^{0.9} \approx 1229.8$.
9. From Example 3, $x^2 + y^2 = C$. At $(0, 4)$: $C = 16$ (the point gives $y = 4$ at $x = 0$). The curve is the upper half of the circle, $y = \sqrt{16 - x^2}$.

An equation $\dfrac{dy}{dx} = f(x, y)$ is homogeneous when $f$ depends only on the ratio $\dfrac{y}{x}$, that is, $f(\lambda x, \lambda y) = f(x, y)$ (Section 9.4). It is not separable directly, but a change of variable makes it so.

The method. Put $y = v\,x$, where $v$ is a new unknown function of $x$. By the product rule,

$$\frac{dy}{dx} = v + x\,\frac{dv}{dx}.$$

Substitute into the equation. The ratio $\frac{y}{x}$ becomes just $v$, and the result is a separable equation in $v$ and $x$. Solve it, then replace $v$ by $\frac{y}{x}$.

Solve $\dfrac{dy}{dx} = \dfrac{x + y}{x}$. Here $\dfrac{x + y}{x} = 1 + \dfrac{y}{x} = 1 + v$. So $v + x\dfrac{dv}{dx} = 1 + v$, which gives $x\dfrac{dv}{dx} = 1$, that is $dv = \dfrac{dx}{x}$. Integrate: $v = \ln|x| + C$. Replace $v = \frac{y}{x}$:

$$y = x\,(\ln|x| + C).$$

The computer checked $y = x(\ln x + 2)$ against the equation for several $x > 0$.

Solve $\dfrac{dy}{dx} = \dfrac{x^2 + y^2}{xy}$. Divide top and bottom by $x^2$: the right side is $\dfrac{1 + v^2}{v}$. So $v + x\dfrac{dv}{dx} = \dfrac{1 + v^2}{v}$, hence $x\dfrac{dv}{dx} = \dfrac{1 + v^2}{v} - v = \dfrac{1}{v}$. Separate: $v\,dv = \dfrac{dx}{x}$. Integrate: $\dfrac{v^2}{2} = \ln|x| + c$. So $y^2 = x^2\,(2\ln|x| + C)$. For $x > 0$ and $y > 0$:

$$y = x\sqrt{2\ln x + C}.$$

Solve $\dfrac{dy}{dx} = \dfrac{y}{x} + \tan\dfrac{y}{x}$. Then $v + x\dfrac{dv}{dx} = v + \tan v$, so $x\dfrac{dv}{dx} = \tan v$. Separate: $\cot v\,dv = \dfrac{dx}{x}$. Integrate: $\ln|\sin v| = \ln|x| + c$, so $\sin v = Cx$. Hence $\sin\dfrac{y}{x} = Cx$, or $y = x\sin^{-1}(Cx)$.

Solve $x\dfrac{dy}{dx} = y + \sqrt{x^2 + y^2}$ for $x > 0$. Divide by $x$: $\dfrac{dy}{dx} = v + \sqrt{1 + v^2}$. So $v + x\dfrac{dv}{dx} = v + \sqrt{1 + v^2}$, which gives $\dfrac{dv}{\sqrt{1 + v^2}} = \dfrac{dx}{x}$. Integrate: $\sinh^{-1}v = \ln x + c$, which gives $v = \sinh(\ln x + c)$. Writing $K = e^c$ and using $\sinh u = \frac{e^u - e^{-u}}{2}$:

$$y = \frac{K x^2 - \frac{1}{K}}{2}.$$

With $K = 1$ this is $y = \frac{x^2 - 1}{2}$, and with $K = 2$ it is $y = x^2 - \frac{1}{4}$. The computer checked both against the original equation.

Find the solution of $\dfrac{dy}{dx} = \dfrac{x + y}{x}$ with $y(1) = 2$. From Example 1, $y = x(\ln x + C)$. At $x = 1$: $2 = 1\cdot(0 + C)$, so $C = 2$. Then $y(e) = e\,(1 + 2) = 3e \approx 8.1548$. A step-by-step numerical solution of the equation from $x = 1$ to $x = e$ agrees.

1. Solve $\dfrac{dy}{dx} = \dfrac{x - y}{x}$ for $x > 0$.
2. Solve $\dfrac{dy}{dx} = \dfrac{y + 2x}{x}$ for $x > 0$.
3. Solve $\dfrac{dy}{dx} = \dfrac{x^2 + 2y^2}{xy}$ for $x > 0$, $y > 0$.
4. Solve $\dfrac{dy}{dx} = \dfrac{y}{x} + \sin\dfrac{y}{x}$ for $x > 0$.
5. Which of these are homogeneous: (a) $\dfrac{dy}{dx} = \dfrac{x^2 + y^2}{xy}$, (b) $\dfrac{dy}{dx} = \dfrac{x}{y}$, (c) $\dfrac{dy}{dx} = \dfrac{x^2 - y^2}{2xy}$, (d) $\dfrac{dy}{dx} = \dfrac{x^3 + y^3}{x^2y^2}$, (e) $\dfrac{dy}{dx} = \dfrac{x + 3}{y}$?
6. Solve $\dfrac{dy}{dx} = \dfrac{3x + 2y}{x}$ with $y(1) = 0$.
7. Solve $\dfrac{dy}{dx} = \dfrac{y}{x} + \dfrac{y^2}{x^2}$ for $x \ge 1$.
8. Solve $\dfrac{dy}{dx} = \dfrac{x^2 + y^2}{xy}$ with $y(1) = 1$ and find $y(e)$.

Answers

1. $\dfrac{x - y}{x} = 1 - v$, so $v + x\dfrac{dv}{dx} = 1 - v$ and $x\dfrac{dv}{dx} = 1 - 2v$. Separate: $\dfrac{dv}{1 - 2v} = \dfrac{dx}{x}$. Integrate: $-\dfrac{1}{2}\ln|1 - 2v| = \ln x + c$, so $1 - 2v = \dfrac{K}{x^2}$ and $y = \dfrac{x^2 - K}{2x}$. (The computer checked $K = 1$.)
2. $\dfrac{y + 2x}{x} = v + 2$, so $x\dfrac{dv}{dx} = 2$, $v = 2\ln x + C$, and $y = x\,(2\ln x + C)$.
3. The right side is $\dfrac{1 + 2v^2}{v}$, so $x\dfrac{dv}{dx} = \dfrac{1 + v^2}{v}$. Separate: $\dfrac{v\,dv}{1 + v^2} = \dfrac{dx}{x}$. Then $\dfrac{1}{2}\ln(1 + v^2) = \ln x + c$, so $1 + v^2 = Kx^2$ and $y = x\sqrt{Kx^2 - 1}$. (Checked for $K = 2$.)
4. $x\dfrac{dv}{dx} = \sin v$. Separate: $\csc v\,dv = \dfrac{dx}{x}$. Integrate: $\ln\left|\tan\dfrac{v}{2}\right| = \ln x + c$, so $\tan\dfrac{y}{2x} = Cx$ and $y = 2x\tan^{-1}(Cx)$. (Checked for $C = 1$.)
5. (a), (b), (c) are homogeneous: each is unchanged when $x \to \lambda x$, $y \to \lambda y$. (d) is not: the right side becomes $\frac{1}{\lambda}\cdot\frac{x^3 + y^3}{x^2y^2}$ (degree $3$ on top, degree $4$ below). (e) is not: the number $3$ breaks it.
6. The right side is $3 + 2v$, so $x\dfrac{dv}{dx} = 3 + v$. Separate: $\dfrac{dv}{3 + v} = \dfrac{dx}{x}$. Integrate: $3 + v = Kx$, so $y = Kx^2 - 3x$. From $y(1) = K - 3 = 0$: $K = 3$. The solution is $y = 3x^2 - 3x$.
7. $v + x\dfrac{dv}{dx} = v + v^2$, so $\dfrac{dv}{v^2} = \dfrac{dx}{x}$. Then $-\dfrac{1}{v} = \ln x + C$, so $y = -\dfrac{x}{\ln x + C}$. (Checked for $C = 1$.)
8. From Example 2, $y^2 = x^2(2\ln x + C)$. At $(1, 1)$: $1 = C$. So $y = x\sqrt{2\ln x + 1}$ and $y(e) = e\sqrt{3} \approx 4.7082$. A step-by-step numerical solution agrees.