Skip to content

Mathematics · Ch 9 — Differential Equations

Methods of Solving First Order, First Degree Differential Equations

9.4

Methods of Solving First Order, First Degree Differential Equations

Methods of Solving First Order, First Degree Differential Equations

A differential equation of the form dydx=f(x,y)\frac{dy}{dx} = f(x, y) is called first order, first degree: the highest derivative present is the first, and it appears only to the first power. We study three systematic methods, plus Bernoulli's equation, according to the form of f(x,y)f(x, y).

1. Differential Equations with Variables Separable

If f(x,y)=g(x) h(y)f(x, y) = g(x) \, h(y) — a product of a function of xx and a function of yy — the equation has separable variables:

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

Method of solution:

  1. Separate the variables — all yy terms (with dydy) on one side, all xx terms (with dxdx) on the other:

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

  1. Integrate both sides:

∫1h(y) dy=∫g(x) dx+C\int \frac{1}{h(y)} \, dy = \int g(x) \, dx + C

  1. The resulting relation between xx and yy (involving CC) is the general solution.
Watch out

Dividing by h(y)h(y) assumes h(y)≠0h(y) \neq 0. Constant solutions y=y0y = y_0 with h(y0)=0h(y_0) = 0 must be checked separately; they are often singular solutions not obtainable from the general solution for any finite CC.

Example: Solve dydx=1+y21+x2\frac{dy}{dx} = \frac{1 + y^2}{1 + x^2}. Separating and integrating:

dy1+y2=dx1+x2  ⇒  tan⁡−1y=tan⁡−1x+C\frac{dy}{1 + y^2} = \frac{dx}{1 + x^2} \;\Rightarrow\; \tan^{-1} y = \tan^{-1} x + C

Using tan⁡−1a−tan⁡−1b=tan⁡−1(a−b1+ab)\tan^{-1} a - \tan^{-1} b = \tan^{-1}\left(\frac{a-b}{1+ab}\right), this can be written as y−x1+xy=k\frac{y-x}{1+xy} = k, i.e. y−x=k(1+xy)y - x = k(1 + xy), an implicit form of the solution.

2. Homogeneous Differential Equations

F(x,y)F(x, y) is homogeneous of degree nn if F(λx,λy)=λnF(x,y)F(\lambda x, \lambda y) = \lambda^n F(x, y) for any non-zero λ\lambda. The equation

dydx=P(x,y)Q(x,y)\frac{dy}{dx} = \frac{P(x, y)}{Q(x, y)}

is homogeneous when P(x,y)P(x, y) and Q(x,y)Q(x, y) are homogeneous of the same degree.

Note

Equivalently, it is homogeneous if it can be written dydx=F(yx)\frac{dy}{dx} = F\left(\frac{y}{x}\right) — the right-hand side depends only on the ratio y/xy/x.

Method of solution (substitution y=vxy = vx):

  1. Write the equation as dydx=F(yx)\frac{dy}{dx} = F\left(\frac{y}{x}\right).
  2. Substitute y=vxy = vx, so dydx=v+xdvdx\frac{dy}{dx} = v + x \frac{dv}{dx}.
  3. The equation becomes v+xdvdx=F(v)v + x \frac{dv}{dx} = F(v), which separates to

dvF(v)−v=dxx\frac{dv}{F(v) - v} = \frac{dx}{x}

  1. Integrate both sides and replace vv by y/xy/x to obtain the general solution.
Important

The substitution y=vxy = vx always reduces a homogeneous equation to a separable one. The constant solution where F(v0)=v0F(v_0) = v_0 (giving y=v0xy = v_0 x, a straight line through the origin) must be checked separately.

Example: Solve (x2+y2) dx−2xy dy=0(x^2 + y^2) \, dx - 2xy \, dy = 0.

dydx=x2+y22xy=1+(y/x)22(y/x)=F(yx)\frac{dy}{dx} = \frac{x^2 + y^2}{2xy} = \frac{1 + (y/x)^2}{2(y/x)} = F\left(\frac{y}{x}\right)

Substitute y=vxy = vx, dydx=v+xdvdx\frac{dy}{dx} = v + x\frac{dv}{dx}:

xdvdx=1+v22v−v=1−v22v  ⇒  2v1−v2 dv=dxxx\frac{dv}{dx} = \frac{1 + v^2}{2v} - v = \frac{1 - v^2}{2v} \;\Rightarrow\; \frac{2v}{1 - v^2} \, dv = \frac{dx}{x}

Integrating:

−ln⁡∣1−v2∣=ln⁡∣x∣+ln⁡∣C∣  ⇒  11−v2=Cx-\ln|1 - v^2| = \ln|x| + \ln|C| \;\Rightarrow\; \frac{1}{1 - v^2} = Cx

Replacing v=y/xv = y/x gives x=C(x2−y2)x = C(x^2 - y^2), the general solution.

3. Linear Differential Equations

A first order linear differential equation has the form

dydx+P(x) y=Q(x)\frac{dy}{dx} + P(x) \, y = Q(x)

where P(x)P(x) and Q(x)Q(x) are functions of xx only (or constants).

Note

The equation is "linear" because yy and dydx\frac{dy}{dx} appear only to the first power and are not multiplied together. If Q(x)=0Q(x) = 0 it is homogeneous linear; otherwise non-homogeneous linear.

Method of solution (integrating factor): multiply through by a function μ(x)\mu(x) that makes the left-hand side an exact derivative.

  1. In standard form, compute the integrating factor

I.F.=e∫P(x) dx\text{I.F.} = e^{\int P(x) \, dx}

  1. Multiplying through, the left-hand side becomes ddx(y e∫P dx)\frac{d}{dx}\left( y \, e^{\int P \, dx} \right), so

ddx(y e∫P dx)=Q e∫P dx\frac{d}{dx} \left( y \, e^{\int P \, dx} \right) = Q \, e^{\int P \, dx}

  1. Integrate and solve for yy:

y e∫P dx=∫Q e∫P dx dx+C  ⇒  y=e−∫P dx[∫Q e∫P dx dx+C]y \, e^{\int P \, dx} = \int Q \, e^{\int P \, dx} \, dx + C \;\Rightarrow\; y = e^{-\int P \, dx} \left[ \int Q \, e^{\int P \, dx} \, dx + C \right]

›Proof

Derivation of the integrating factor. We want μ(x)\mu(x) with μdydx+μPy=ddx(μy)=μdydx+ydμdx\mu \frac{dy}{dx} + \mu P y = \frac{d}{dx}(\mu y) = \mu \frac{dy}{dx} + y \frac{d\mu}{dx}. Cancelling μdydx\mu \frac{dy}{dx} gives μPy=ydμdx\mu P y = y \frac{d\mu}{dx}, i.e. dμμ=P dx\frac{d\mu}{\mu} = P \, dx. Integrating, ln⁡∣μ∣=∫P dx\ln|\mu| = \int P \, dx, so μ=e∫P dx\mu = e^{\int P \, dx}.

Example: Solve dydx+2y=e3x\frac{dy}{dx} + 2y = e^{3x}. Here P=2P = 2, Q=e3xQ = e^{3x}, so I.F. =e2x= e^{2x}:

ddx(ye2x)=e5x  ⇒  ye2x=e5x5+C\frac{d}{dx} \left( y e^{2x} \right) = e^{5x} \;\Rightarrow\; y e^{2x} = \frac{e^{5x}}{5} + C

y=e3x5+Ce−2xy = \frac{e^{3x}}{5} + C e^{-2x}

4. Equations Reducible to Linear Form (Bernoulli's Equation)

An equation

dydx+P(x) y=Q(x) yn(n≠0,1)\frac{dy}{dx} + P(x) \, y = Q(x) \, y^n \qquad (n \neq 0, 1)

is called Bernoulli's equation. It is not linear, but reduces to linear form by a substitution.

Method of solution:

  1. Divide by yny^n (assuming y≠0y \neq 0):   y−ndydx+P(x) y1−n=Q(x)\; y^{-n} \frac{dy}{dx} + P(x) \, y^{1-n} = Q(x). …
Start your 14-day free trial to unlock the full solution →
← Previous · 9.3Q12 — The number of arbitrary constants in the particular solution of a diff…Next · 9.4.1 →Differential Equations with Variables Separable