Exact & substitutions practice problems

You solve equations written as M dx + N dy = 0. Check one test: does M_y equal N_x? If yes, find a potential function F and write F(x, y) = C. If no, multiply by an integrating factor first. Substitutions v = y/x and u = y^(1-n) turn two other forms into earlier methods. Recognition cue: the equation arrives as M dx + N dy = 0, the right side depends only on y/x, or one power y^n spoils a linear equation. You will practice: exactness test, potential function, integrating factor, homogeneous substitution, Bernoulli substitution.

10 problems, each with a worked answer. Work them on paper, in order — the difficulty ramps gently.

Let M = 2xy and N = x². Compute M_y, the derivative of M with respect to y, and N_x, the derivative of N with respect to x. Is the equation M dx + N dy = 0 exact?

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Answer: $M_{y} = 2 x, N_{x} = 2 x$

Differentiate M with respect to y and N with respect to x. The results are equal, so the equation is exact.

M_{y} = 2 x, N_{x} = 2 x

Solve (2xy) dx + (x²) dy = 0. Give the general solution.

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Answer: $x^{2} y = C$

The equation is exact. The potential function is F = x²y.

F = x^{2} y, x^{2} y = C

The equation dy/dx + y = y³ is a Bernoulli equation with n = 3. Which substitution u turns it into a linear equation?

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Answer: $u = y^{- 2}$

Bernoulli uses u = y^(1-n). Here n = 3.

u = y^{1 - 3} = y^{- 2}

Solve y dx + x dy = 0 with y(2) = 3.

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Answer: $y = \frac{6}{x}$

The equation is exact with potential F = xy. Apply the initial condition.

x y = C, C = 2 \cdot 3 = 6

y = \frac{6}{x}

Solve (2x + y) dx + (x + 2y) dy = 0. Give the general solution.

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Answer: $x^{2} + x y + y^{2} = C$

M_y = 1 = N_x, so the equation is exact. Integrate M in x, then match F_y to N.

F = x^{2} + x y + g (y), g^{'} (y) = 2 y

x^{2} + x y + y^{2} = C

Solve (y cos x) dx + (sin x) dy = 0 with y(π/2) = 2.

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Answer: $y sin x = 2$

M_y = cos x = N_x, so the equation is exact with F = y sin x. Insert the initial point.

y sin x = C, C = 2 \cdot 1 = 2

The equation (3y) dx + (x) dy = 0 is not exact. Find an integrating factor of the form μ(x).

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Answer: $μ = x^{2}$

Compute the ratio (M_y - N_x)/N. It depends on x only.

\frac{M _{y} - N _{x}}{N} = \frac{3 - 1}{x} = \frac{2}{x}

μ = e^{\int \frac{2}{x} d x} = x^{2}

Solve y dx + 2x dy = 0. Use an integrating factor of the form μ(y). Give the general solution.

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Answer: $x y^{2} = C$

Compute (N_x - M_y)/M = 1/y, so μ = y. Multiply and solve the exact equation.

μ = y, y^{2} d x + 2 x y d y = 0

x y^{2} = C

Solve x dy/dx = x + y with y(1) = 0.

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Answer: $y = x ln x$

Divide by x to get dy/dx = 1 + y/x. Substitute v = y/x.

v + x \frac{d v}{d x} = 1 + v, x \frac{d v}{d x} = 1

v = ln x + C, y = x ln x + C x

The initial condition y(1) = 0 gives C = 0.

10.

A product's market share s(t) satisfies ds/dt = s² - s with s(0) = 1/2. Find s(t).

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Answer: $s = \frac{1}{1 + e ^{t}}$

This is Bernoulli with n = 2. Set u = 1/s.

\frac{d u}{d t} = u - 1, u = 1 + C e^{t}

u (0) = 2, C = 1, s = \frac{1}{1 + e ^{t}}

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Exact & substitutions practice problems

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