Separable equations practice problems

You solve equations of the form dy/dx = g(x)·h(y). Move every y term to one side and every x term to the other, integrate both sides, then solve for y. An initial condition fixes the constant C. Recognition cue: the right side of dy/dx factors into a function of x alone times a function of y alone. You will practice: Split and integrate, Use the initial condition, Check lost constant solutions, Interval of validity.

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

Find the general solution of dy/dx = 2y.

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

Separate, integrate, then exponentiate.

\frac{d y}{y} = 2 d x, ln ∣ y ∣ = 2 x + C

y = C e^{2 x}

Find the general solution of dy/dx = 3x².

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Answer: $y = x^{3} + C$

The right side has no y, so integrate directly.

y = \int 3 x^{2} d x = x^{3} + C

Solve dy/dx = -y with y(0) = 5.

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Answer: $y = 5 e^{- x}$

Separate and integrate, then fit C.

\frac{d y}{y} = - d x, ln ∣ y ∣ = - x + C, y = C e^{- x}

5 = C e^{0}, C = 5, y = 5 e^{- x}

Find the general solution of dy/dx = xy.

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

Separate and integrate, then exponentiate.

\frac{d y}{y} = x d x, ln ∣ y ∣ = \frac{x ^{2}}{2} + C

y = C e^{x^{2} /2}

Solve dy/dx = y² with y(1) = 1.

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Answer: $y = \frac{1}{2 - x}$

Separate, integrate, then use the condition at x = 1.

\frac{d y}{y ^{2}} = d x, - \frac{1}{y} = x + C

- 1 = 1 + C, C = - 2, y = \frac{1}{2 - x}

A population grows at dP/dt = 0.02P with P(0) = 1000. Find P(t).

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Answer: $P = 1000 e^{0.02 t}$

Separate and integrate, then fit C with P(0) = 1000.

\frac{d P}{P} = 0.02 d t, P = C e^{0.02 t}

C = 1000, P = 1000 e^{0.02 t}

Solve y · dy/dx = e^x with y(0) = 2. Give y explicitly.

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

The equation is already separated. Integrate, fit C, then take the positive root because y(0) = 2 is positive.

y d y = e^{x} d x, \frac{y ^{2}}{2} = e^{x} + C

2 = 1 + C, C = 1, y^{2} = 2 e^{x} + 2, y = 2 e^{x} + 2

Find the general solution of dy/dx = 6xy². Also list any constant solution.

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Answer: $y = \frac{- 1}{3 x ^{2} + C}, y = 0$

Separate and integrate.

\frac{d y}{y ^{2}} = 6 x d x, - \frac{1}{y} = 3 x^{2} + C, y = \frac{- 1}{3 x ^{2} + C}

Dividing by y² lost the constant solution y = 0. Add it back.

A radioactive sample decays at dN/dt = -0.5N with N(0) = 80. Find N(t).

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Answer: $N = 80 e^{- 0.5 t}$

Separate and integrate, then fit C with N(0) = 80.

\frac{d N}{N} = - 0.5 d t, N = C e^{- 0.5 t}

C = 80, N = 80 e^{- 0.5 t}

10.

Find the general solution of dy/dx = y cos x.

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Answer: $y = C e^{s i n x}$

Separate and integrate, then exponentiate.

\frac{d y}{y} = cos x d x, ln ∣ y ∣ = sin x + C

y = C e^{s i n x}

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Separable equations practice problems

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