Differential equations

First-order differential equations

The key to first-order equations is recognizing the type and choosing the method. Find the general solution, then use the initial condition to fix the constant.

Basic, Standard: University Year 1 · 2nd semester / Advanced: University Year 2+

BasicStandardAdvancedMix

0Make a worksheet on this topic

Separable equations

Separable

\frac{dy}{dx} = f(x)g(y) \quad\Rightarrow\quad \int \frac{dy}{g(y)} = \int f(x)\,dx

Example: for $y' = 2xy$ , $\int \frac{dy}{y} = \int 2x\,dx$ gives $\log|y| = x^2 + C_1$ , so $y = Ce^{x^2}$ . With $y(0) = 3$ , $C = 3$ .

Homogeneous equations

For $y' = f\left(\frac{y}{x}\right)$ , put $y = ux$ . Then $y' = u + xu'$ and

xu' = f(u) - u,

which is separable. Finally substitute back $u = \frac{y}{x}$ .

Linear first-order equations

Linear

y' + P(x)y = Q(x) \quad\Rightarrow\quad y = e^{-\int P\,dx}\left(\int Q e^{\int P\,dx}\,dx + C\right)

Multiplying by the integrating factor $e^{\int P\,dx}$ turns the left side into $\left(e^{\int P\,dx}y\right)'$ . With constant coefficients it is quicker to add a particular solution to the homogeneous solution $Ce^{-px}$ .

A Bernoulli equation $y' + Py = Qy^n$ becomes linear with $z = y^{1-n}$ . For an exact equation $P\,dx + Q\,dy = 0$ (with $P_y = Q_x$ ), find $F$ with $F_x = P$ , $F_y = Q$ ; the solutions are $F(x,\ y) = C$ .

Q1BasicUniv Y1 · Sem 2First-order differential equations

Find the solution of the differential equation that satisfies $y(0) = \log 5$ .

y' = e^{x - y}

Answer

y = \log(e^{x} + 4)

Solution

Separating variables and integrating,
$\int e^{y}\,dy = \int e^{x}\,dx \quad\Rightarrow\quad e^{y} = e^{x} + C$
From $y(0) = \log 5$ , $5 = 1 + C$ , so $C = 4$ and
$y = \log(e^{x} + 4)$

Q2StandardUniv Y1 · Sem 2First-order differential equations

Find the solution of the differential equation that satisfies $y(0) = -3$ .

y' + 3y = 9x - 6

Answer

y = 3x - 3

Solution

The homogeneous equation $y' + 3y = 0$ has general solution
$y = Ce^{-3x}$
Try $y = ax + b$ ; substituting and comparing coefficients gives $a = 3$ , $b = -3$ .
$y_p = 3x - 3$
Substituting $y(0) = -3$ into $y = 3x - 3 + Ce^{-3x}$ gives $C = 0$ , so
$y = 3x - 3$

Q3AdvancedUniv Y2+First-order differential equations

Find the solution of the differential equation that satisfies $y(0) = 0$ .

y' - y = \cos x

Answer

y = \frac{1}{2}\sin x - \frac{1}{2}\cos x + \frac{1}{2}e^{x}

Solution

Try $y = A\sin x + B\cos x$ ; comparing coefficients,
$A = \frac{1}{2},\ B = -\frac{1}{2},\quad y_p = \frac{1}{2}\sin x - \frac{1}{2}\cos x$
The general solution is $y = y_p + Ce^{x}$ ; $y(0) = 0$ gives $C = \frac{1}{2}$ , so
$y = \frac{1}{2}\sin x - \frac{1}{2}\cos x + \frac{1}{2}e^{x}$

Q1BasicUniv Y1 · Sem 2First-order differential equations

Find the solution of the differential equation that satisfies $y(0) = \log 3$ .

y' = e^{x - y}

Q2StandardUniv Y1 · Sem 2First-order differential equations

Find the solution of the differential equation that satisfies $y(1) = 3$ .

y' + \frac{2y}{x} = x\quad (x > 0)

Q3AdvancedUniv Y2+First-order differential equations

Find the solution of the differential equation that satisfies $y(0) = 0$ .

y' + y = \sin x

Teleport

Main features

On this page

Topics

Display

First-order differential equations

Separable equations

Homogeneous equations

Linear first-order equations

Hint

Answer

Solution

Hint

Answer

Solution

Hint

Answer

Solution

Hint

Answer

Solution

Hint

Answer

Solution

Hint

Answer

Solution

Math problem generator

Separable equations

Homogeneous equations

Linear first-order equations

Worked examples

Hint

Answer

Solution

Hint

Answer

Solution

Hint

Answer

Solution

Practice problems

Hint

Answer

Solution

Hint

Answer

Solution

Hint

Answer

Solution

Related topics

Second-order linear ODEs with constant coefficients

Basic indefinite integrals

Integration by substitution