← Differential Equations

Unit 2: Solving First-Order Equations

The toolbox for a first-order equation: separating variables, integrating factors, exactness, and the substitutions that turn a hard equation into an easy one.

7 topics · 29 questions

Each topic is a path, not a menu: work down it in order and every question builds on the one above. Click any of them to watch the answer draw itself.

2.1 Separable Equations

  1. How it worksHow does rewriting dy/dx as a ratio let you separate a differential equation into two ordinary integrals?
  2. Worked exampleSolve the separable equation dy/dx = xy given that y(0) = 2.
  3. Worked exampleSolve the separable equation y' = y2 cos(x).
  4. Common mistakeWhy can dividing both sides of a separable equation by y quietly delete the solution y = 0?
  5. Going furtherWhat happens when a separable equation's variables can't be fully split apart because of a stray cross term?
Your turn1 of 2
Solve the initial value problem dy/dx = 3xy, y(0) = 5, for y > 0. Find y(x).

2.2 Linear First-Order Equations and the Integrating Factor

  1. How it worksHow do you rewrite a first-order linear equation into the standard form y' + p(x)y = g(x)?
  2. How it worksHow does the integrating factor e to the integral of p(x) dx turn the left side into the derivative of a product?
  3. Worked exampleSolve the linear equation y' + 2y = e−x using an integrating factor.
  4. Common mistakeWhy doesn't dropping the constant of integration inside the integrating factor's exponent break the method?
Your turn1 of 2
Find the integrating factor mu(x) for the linear equation y' + 5y = g(x).

2.3 Exact Equations

  1. How it worksHow does checking whether the partial derivative of M with respect to y equals the partial derivative of N with respect to x confirm an equation is exact?
  2. Worked exampleSolve the exact equation (2xy + 3) dx + (x2 - 1) dy = 0.
  3. Common mistakeWhy isn't every equation you can write in the form M dx + N dy = 0 actually exact?
  4. Going furtherWhat happens when an exact equation's implicit solution can't be solved explicitly for y in terms of x?
Your turn1 of 2
For the equation (3xy + 3y2) dx + (4x2 + 2x) dy = 0, with M = 3xy + 3y2 and N = 4x2 + 2x, find (dM/dy) minus (dN/dx) at the point (x, y) = (1, 2). (A value of 0 would mean the equation is exact there.)

2.4 Turning a Non-Exact Equation into an Exact One

  1. How it worksHow do you find an integrating factor depending on x alone from the mismatch between M sub y and N sub x?
  2. Worked exampleFind an integrating factor for (3xy + y2) dx + (x2 + xy) dy = 0 and solve the equation.
  3. Going furtherWhat happens when no integrating factor depending on x alone or y alone exists for an equation?
  4. Common mistakeWhy doesn't multiplying an equation by just any function you like turn it into an exact one?
Your turn1 of 2
For an equation where (M sub y - N sub x)/N = 3/x, the integrating factor is mu(x) = exp(integral of 3/x dx). Find the value of mu(x) at x = 4.

2.5 Bernoulli Equations

  1. How it worksHow does the substitution v = y1-n turn a Bernoulli equation into a linear equation in v?
  2. Worked exampleSolve the Bernoulli equation y' + y = xy3 using the substitution v = y−2.
  3. Common mistakeWhy isn't a Bernoulli equation solved by treating the yn term as if it were just another linear term?
  4. Going furtherWhat happens to the Bernoulli substitution v = y1-n when n equals 0 or n equals 1?
Your turn1 of 2
For the Bernoulli equation y' + 3y = qy3, applying the substitution v = y1-n gives the linear equation v' + 3(1-n)v = q(1-n). Find the coefficient of v in this new equation.

2.6 Homogeneous Equations and the v = y/x Substitution

  1. How it worksHow does the substitution v = y/x turn a homogeneous equation into a separable one?
  2. Worked exampleSolve the homogeneous equation dy/dx = (x + y)/x using the substitution v = y/x.
  3. Common mistakeWhy isn't a homogeneous equation here the same idea as a homogeneous linear equation with zero forcing?
  4. Going furtherWhat happens to the substitution when the equation is written in terms of x/y instead of y/x?
Your turn1 of 2
Solve the initial value problem dy/dx = 2 + y/x, y(4) = 9, for x > 0, using the substitution v = y/x. Find y(x).

2.7 Choosing Which First-Order Method to Use

  1. How it worksHow do you decide in order whether an equation is separable, linear, exact, or needs a substitution?
  2. Worked exampleSolve dy/dx = (2x - y)/x by first figuring out which method actually applies to it.
  3. Common mistakeWhy isn't there one universal method that solves every first-order equation you'll ever encounter?
  4. Ties togetherHow does recognizing an equation's type here connect to matching a technique in an integral calculus course?
Your turn1 of 2
Solve the initial value problem dy/dx = (5x - y)/x, y(2) = 5, for x > 0. Find y(x).