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Notes

First Order Differential Equations

Separable Equations

$\frac{dy}{dx} = g(x)h(y)$

To solve:

  1. Rearrange: \(\frac{1}{h(y)}dy = g(x)dx\)

  2. Integrate both sides: \(\int \frac{1}{h(y)}dy = \int g(x)dx\)

  3. Solve for $y$

  4. Lump all constants together into $c_1$

Linear Equations

$a_1(x)\frac{dy}{dx} + a_0(x)y = g(x)$

If $g(x) = 0$, then the equation is said to be homogeneous, otherwise its non-homogeneous.

To solve:

  1. Put into standard form: \(\frac{dy}{dx} + \frac{a_0(x)}{a_1(x)}y = \frac{g(x)}{a_1(x)}\)

  2. $P(x) = \frac{a_0(x)}{a_1(x)}$ and $f(x) = \frac{g(x)}{a_1(x)}$

  3. Determine the integrating factor: $\mu(x) = e^{\int P(x)dx}$
    Note: Ignore the constant when integrating $P(x)$.

  4. Multiply both sides of the equation by the integrating factor: \(\mu(x)[\frac{dy}{dx} + P(x)y] = \mu(x)f(x)\)

  5. Notice that the left side of the equation will always become $(\mu(x)y)’$: \((\mu(x)y)' = \mu(x)f(x)\)

  6. Integrate both sides: \(\mu(x)y = \int \mu(x)f(x)dx\)

  7. Solve for $y$

  8. Lump all constants together into $c_1$

Exact Equations

$M(x, y)dx + N(x, y)dy = 0$

If $f(x, y)$ exists such that $df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy = M(x, y)dx + N(x, y)dy$, then $df = 0$ and $f(x, y) = c$

To solve:

  1. Verify that $M_y = N_x$ or $M_x = N_y$ (whichever is easier): \(\frac{\partial}{\partial y}M(x, y) = \frac{\partial}{\partial x}N(x, y)\)

  2. Integrate $M(x, y)$. The constant of integration will be replaced by a function that only depends on $y$: \(\int M(x, y)dx = \mathbf{M}(x, y) + h(y)\)

  3. Integrate $N(x, y)$. The constant of integration will be replaced by a function that only depends on $x$: \(\int N(x, y)dy = \mathbf{N}(x, y) + g(x)\)

  4. $h(y)$ is equal to the terms in $\mathbf{N}(x, y)$ that aren’t also in $\mathbf{M}(x, y)$.

  5. $g(y)$ is equal to the terms in $\mathbf{M}(x, y)$ that aren’t also in $\mathbf{N}(x, y)$.

  6. Determine $f(x, y)$. Terms that exist in both $\mathbf{M}(x, y)$ and $\mathbf{N}(x, y)$ exist only once in $f(x, y)$. Also include $h(y)$ and $g(y)$ in $f(x, y)$.

  7. Rewrite as $f(x, y) = c$ to get an implicit solution.

Bernoulli’s Equation

$a_1(x)\frac{dy}{dx} + a_0(x)y = g(x)y^n$

If $n = 0$ or $1$, then the equation becomes linear and can be solved using the method above.

To solve:

  1. Rewrite in standard form: \(\frac{dy}{dx} + \frac{a_0(x)}{a_1(x)}y = \frac{g(x)}{a_1(x)}y^n\)

  2. $P(x) = \frac{a_0(x)}{a_1(x)}$ and $f(x) = \frac{g(x)}{a_1(x)}$

  3. Compute $u = y^{n - 1}$. Solve in terms of $y$, then take the derivative to get $\frac{dy}{dx}$: \(u = y^{n - 1} \Rightarrow y = u^{(1 - n)}\) \(\frac{dy}{dx} = (1 - n)u^{-n}\)

  4. Substitute in $y$ and $y’$: \((1 - n)u^{-n} + P(x)u^{(1 - n)} = f(x)(u^{(1 - n)})^n\)

  5. Solve as a linear equation.

  6. Substitute back in for $u$ and $u’$.

Reduction to Separable

$y’ = f(Ax + By + C)$

To solve:

  1. Find a substitution of the form $u = Ax + By + C$

  2. Solve $u$ in terms of $y$ and $y’$: \(u = Ax + By + C \Rightarrow By = u - Ax - C\) \(y = \frac{1}{B}u - \frac{A}{B}x - \frac{C}{B}\) \(y' = \frac{1}{B}u' - \frac{A}{B}\)

  3. Substitute in $y$ and $y’$

  4. Solve as a separable equation.

  5. Substitute back in for $u$

Homogeneous Equations

Not to be confused with homogeneous linear equations when $f(x) = 0$.

Any equation (not just a DE) is homogeneous to the degree $\alpha$ if: $f(tx, ty) = t^{\alpha}f(x, y)$

$M(x, y)dx + N(x, y)dy = 0$

If the equation has the form of an exact equation, but $M_y = N_x$, this method can be used if $M(x, y)$ and $N(x, y)$ are homogeneous to the same degree.

To solve:

  1. Substitute $y = ux$ and $y’ = udx + xdu$ or $x = vy$ and $x’ = vdy + ydv$, whichever is easier.

  2. Solve as a separable equation.

  3. Substitute back in for $u$ or $v$.