Differential Equations

Problem. State the order and degree of each equation.

(a) \dfrac{dy}{dx} + 3y = e^{x} (b) \dfrac{d^{2}y}{dx^{2}} + 5\left(\dfrac{dy}{dx}\right)^{3} - 4y = 0 (c) \left(\dfrac{d^{3}y}{dx^{3}}\right)^{2} + \dfrac{dy}{dx} = \sin x (d) \dfrac{d^{2}y}{dx^{2}} = \sqrt{1+\left(\dfrac{dy}{dx}\right)^{2}}

Solution.

(a) The highest derivative is \frac{dy}{dx}, appearing to the first power. Order 1, degree 1.

(b) The highest derivative is \frac{d^{2}y}{dx^{2}}, to the first power. The cube belongs to a lower derivative and does not affect the degree. Order 2, degree 1.

(c) Highest derivative \frac{d^{3}y}{dx^{3}}, squared. Order 3, degree 2.

(d) Squaring gives \left(\frac{d^{2}y}{dx^{2}}\right)^{2} = 1+\left(\frac{dy}{dx}\right)^{2}, so the highest derivative is of order 2 and now appears squared. Order 2, degree 2.

Problem. State the order and degree of (a) x\left(\dfrac{d^{2}y}{dx^{2}}\right)^{3} + y\left(\dfrac{dy}{dx}\right)^{4} + y^{5} = 0, (b) \dfrac{d^{2}y}{dx^{2}} + \sin\left(\dfrac{dy}{dx}\right) = 0, (c) \left(\dfrac{dy}{dx}\right)^{3} - 4\dfrac{dy}{dx} + y = 0.

Solution.

(a) Highest derivative: order 2, and it is cubed. Order 2, degree 3. The powers 4 and 5 are irrelevant to both answers.

(b) The highest derivative is of order 2 and appears to the first power, so the order is 2. But the equation is not polynomial in \frac{dy}{dx} — \sin of a derivative cannot be cleared — so the degree is not defined.

(c) Order 1, and the first derivative is cubed. Order 1, degree 3. Notice that order and degree are quite independent of each other.

Problem. Form the differential equation of (a) the family of straight lines y = mx through the origin, (b) the family of circles x^{2}+y^{2} = a^{2} centred at the origin, (c) the family y = Ae^{2x}.

Solution. Each family has one constant, so differentiate once.

(a) y = mx gives \frac{dy}{dx} = m. From the original equation m = \frac{y}{x}, so eliminating m,

(b) Differentiating x^{2}+y^{2}=a^{2} with respect to x: 2x + 2y\frac{dy}{dx} = 0, and a has already vanished, so

This is the equation whose slope field is drawn in §14.3 — the circles are everywhere perpendicular to the radius, which is exactly what \frac{dy}{dx} = -\frac{x}{y} says.

(c) \frac{dy}{dx} = 2Ae^{2x} = 2y, so

Problem. Verify that y = x - 1 + Ce^{-x} is the general solution of \dfrac{dy}{dx} + y = x, and find the solution with y(0) = 2.

Solution. Verification. Differentiate: \frac{dy}{dx} = 1 - Ce^{-x}. Add y:

The Ce^{-x} terms cancel for every C, so the whole family satisfies the equation, which is what "general solution" means.

Particular solution. Put x=0 and y=2 into y = x-1+Ce^{-x}:

That is the red curve in the second figure. Always substitute the condition into the general solution, never into the differential equation.

Problem. Solve \left(1+x^{2}\right)\dfrac{dy}{dx} = 1+y^{2}.

Solution. Separate, putting everything in y on the left:

Integrate both sides, using the standard form \int\frac{dt}{1+t^{2}} = \tan^{-1}t:

That is the general solution. It can also be written \tan^{-1}y - \tan^{-1}x = C, or, taking the tangent of both sides and using the subtraction formula, \frac{y-x}{1+xy} = k — a family of circles.

Check. Differentiating \tan^{-1}y - \tan^{-1}x = C gives \frac{1}{1+y^{2}}\frac{dy}{dx} - \frac{1}{1+x^{2}} = 0, which rearranges to the original equation.

Problem. Solve \dfrac{dy}{dx} = e^{x-y}.

Solution. The exponential factorises, which is what makes this separable:

Integrating both sides,

or y = \ln\left(e^{x}+C\right). Note that e^{x-y} had to be split before the variables could be separated — always look for a hidden product.

Problem. Solve \sec^{2}x\tan y\,dx + \sec^{2}y\tan x\,dy = 0.

Solution. Divide throughout by \tan x\tan y, which separates the variables:

Each term is now \frac{g'}{g}, since \frac{d}{dx}\tan x = \sec^{2}x:

Combining the logarithms and removing them,

Writing the constant as \ln|c| rather than C at the start is a standard trick whenever the answer is going to be a product.

Problem. Solve x\left(1+y^{2}\right)dx = y\left(1+x^{2}\right)dy and find the particular solution through (0,1).

Solution. Divide both sides by \left(1+x^{2}\right)\left(1+y^{2}\right):

Both sides are \frac{g'}{2g} patterns, so integrating gives

Multiply by 2 and remove the logarithms:

Particular solution. Put x=0, y=1: \;1 = c(2), so c = \frac12, giving 2\left(1+x^{2}\right) = 1+y^{2}, that is y^{2} = 2x^{2}+1.

Problem. Solve x\dfrac{dy}{dx} = y(x-1).

Solution. Separate:

Integrate both sides:

Collect the logarithms: \ln|y| + \ln|x| = x + \ln|C|, so \ln|xy| - \ln|C| = x and therefore

Check. \frac{dy}{dx} = \frac{Ce^{x}(x-1)}{x^{2}} = \frac{y(x-1)}{x}, as required.

Problem. Solve \dfrac{dy}{dx} = \dfrac{x^{2}+y^{2}}{2xy}.

Solution. Numerator and denominator are both homogeneous of degree 2, so put y = vx, \frac{dy}{dx} = v + x\frac{dv}{dx}:

Separate the variables:

The left side is -\frac{g'}{g} with g = 1-v^{2}, so integrating gives

Finally put v = \frac yx:

Check. Differentiating x^{2}-y^{2}=Cx gives 2x - 2y\frac{dy}{dx} = C = \frac{x^{2}-y^{2}}{x}, and solving for \frac{dy}{dx} returns \frac{x^{2}+y^{2}}{2xy}.

Problem. Solve \dfrac{dy}{dx} = \dfrac{x-y}{x+y}.

Solution. Both parts are homogeneous of degree 1. Put y = vx:

Separating,

The derivative of 1-2v-v^{2} is -2-2v = -2(1+v), so the left side is -\frac{1}{2}\cdot\frac{g'}{g}:

Multiply by -2 and combine:

With v = \frac yx this is

Problem. Solve x\dfrac{dy}{dx} = y + x\tan\dfrac{y}{x}.

Solution. Divide through by x to expose the ratio:

Put y = vx:

The v terms cancel, which always happens when the equation has this shape:

Problem. Solve \dfrac{dy}{dx} = \dfrac{x+2y-3}{2x+y-3}.

Solution. Step 1 — find the shift. Solve x+2y-3 = 0 and 2x+y-3 = 0 simultaneously. Subtracting gives y = x, and then 3x = 3, so x = 1, y = 1. Put

Step 2 — the equation is now homogeneous.

Step 3 — put Y = vX.

Step 4 — separate and integrate.

Partial fractions: \frac{2+v}{(1-v)(1+v)} = \frac{A}{1-v} + \frac{B}{1+v} gives 2+v = A(1+v) + B(1-v). Put v=1: 3 = 2A, so A = \frac32. Put v=-1: 1 = 2B, so B = \frac12. Hence

Multiply by 2 and collect:

Step 5 — go back. With v = \frac YX, the left side is \frac{(X+Y)X^{2}}{(X-Y)^{3}}, so the X^{2} cancels and

Finally X = x-1 and Y = y-1, so X+Y = x+y-2 and X-Y = x-y:

Problem. Solve x\dfrac{dy}{dx} + 2y = x^{3}.

Solution. Step 1. Divide by x to reach standard form:

Step 2. \int P\,dx = \int\frac{2}{x}dx = 2\ln x = \ln x^{2}, so

Step 3.

Check. \frac{dy}{dx} = \frac{3x^{2}}{5} - \frac{2C}{x^{3}}, so x\frac{dy}{dx} + 2y = \frac{3x^{3}}{5} - \frac{2C}{x^{2}} + \frac{2x^{3}}{5} + \frac{2C}{x^{2}} = x^{3}. Correct for every C.

Problem. Solve \dfrac{dy}{dx} + y\cot x = 2\cos x.

Solution. Already in standard form with P = \cot x and Q = 2\cos x.

Multiplying through,

The right side is the derivative of \sin^{2}x, so

where \csc x is the cosecant. Equations with \cot x or \tan x as P always give a trigonometric integrating factor: \sin x and \sec x respectively.

Problem. Solve \left(1+x^{2}\right)\dfrac{dy}{dx} + 2xy = 4x^{2}.

Solution. Divide by 1+x^{2}:

Notice that I is exactly the factor we divided by — that is common, and it is worth spotting, because then the equation was already in the form \frac{d}{dx}(Iy) = \text{something}:

Problem. Solve (a) \dfrac{dy}{dx} + y = e^{-x}, (b) \dfrac{dy}{dx} - y\tan x = \sec x, (c) \dfrac{dy}{dx} + y = e^{x} with y(0)=1.

Solution. (a) P=1, so I = e^{x} and

(b) P = -\tan x, so \int P\,dx = \ln|\cos x| and I = \cos x:

(c) I = e^{x} again, but now Q = e^{x}:

Now apply y(0) = 1: \;1 = \frac12 + C, so C = \frac12 and

Problem. Solve \left(3x^{2}+2y\right)dx + \left(2x+3y^{2}\right)dy = 0.

Solution. Test. M = 3x^{2}+2y and N = 2x+3y^{2}, so

They agree, so the equation is exact.

Find u. Integrate M with respect to x, treating y as a constant:

Differentiate with respect to y and compare with N:

Solution.

Check. The total differential is \left(3x^{2}+2y\right)dx + \left(2x+3y^{2}\right)dy = 0, the original equation.

Problem. Solve \left(2xy+y^{2}\right)dx + \left(x^{2}+2xy\right)dy = 0.

Solution. \frac{\partial M}{\partial y} = 2x+2y and \frac{\partial N}{\partial x} = 2x+2y, so the equation is exact.

So g is a constant and the solution is

Problem. Solve (a) \left(y\cos x + 1\right)dx + \sin x\,dy = 0, (b) \left(x^{2}+y^{2}\right)dx + 2xy\,dy = 0.

Solution. (a) \frac{\partial M}{\partial y} = \cos x and \frac{\partial N}{\partial x} = \cos x, so it is exact.

(b) \frac{\partial M}{\partial y} = 2y and \frac{\partial N}{\partial x} = 2y, so it is exact.

This equation is also homogeneous of degree 2, so y = vx would solve it as well — a reminder that the types overlap, and the cheapest applicable method is the right one.

Problem. The population of a municipality grows at a rate proportional to the population present. It was 40{,}000 in 2068 BS and 50{,}000 in 2078 BS. What will it be in 2088 BS?

Solution. Let P be the population t years after 2068 BS. Then

using P = 40000 when t = 0. At t = 10, P = 50000:

For 2088 BS, t = 20, and since e^{20k} = \left(e^{10k}\right)^{2}:

Notice that the second step avoided evaluating k at all. Whenever the times are equally spaced, work with the ratio and the arithmetic stays exact.

Problem. A bacterial culture doubles in 3 hours. By what factor has it grown after 12 hours?

Solution. With N = N_{0}e^{kt} and N = 2N_{0} at t=3,

After 12 hours, 12 = 4\times 3, so

The culture is 16 times its original size. Exponential growth doubles in a fixed time, no matter how large the population already is.

Problem. A glass of tea at 80\,^{\circ}C is left in a room at 20\,^{\circ}C. After 10 minutes it has cooled to 50\,^{\circ}C. Find its temperature after a further 10 minutes, and the time at which it reaches 30\,^{\circ}C.

Solution. Here \theta_{s} = 20 and \theta_{0} = 80, so the excess starts at 60:

At t=10, \theta = 50, so the excess is 30:

The excess halves every 10 minutes. At t = 20 the excess is 60\times\frac14 = 15, so

For \theta = 30 the excess is 10:

Problem. Radium has a half-life of 1600 years. What fraction of a sample remains after 3200 years? And if a different substance loses 20\% of its mass in 10 days, find its half-life.

Solution. (a) 3200 = 2\times 1600, so two half-lives have passed and

(b) Losing 20\% leaves 80\%, so e^{-10\lambda} = 0.8 and