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Solution to Clairaut's Differential Equation 📂Odinary Differential Equations

Solution to Clairaut's Differential Equation

Definition

The 1st order nonlinear differential equation below is called Clairaut’s equation.

y=xy′+f(y′) y=xy^\prime+f(y^\prime )

Explanation

The Clairaut differential equation is comparatively easier to solve than other nonlinear differential equations such as the Bernoulli differential equation or the Riccati differential equation.

Solution

Differentiate both sides of the given differential equation y=xy′+f(y′)y=xy^\prime+f(y^\prime ) and then organize.

y′=y′+xy′′+y′′f′(y′)  ⟹  xy′′+y′′f′(y′)=0  ⟹  y′′[x+f′(y′)]=0 \begin{align*} && y^\prime = y^\prime+xy^{\prime \prime} + y^{\prime \prime}f^\prime(y^\prime ) \\ \implies && xy^{\prime \prime} + y^{\prime \prime}f^\prime(y^\prime )=0 \\ \implies && y^{\prime \prime} \left[ x + f^\prime ( y^\prime) \right]=0 \end{align*}

  • Case 1. y′′=0y^{\prime \prime}=0

    In this case, y=ax+by=ax+b can be obtained directly. a, ba,\ b is an arbitrary constant. When substituted into the given differential equation, it results in

    ax+b=xa+f(a)  ⟹  b=f(a) \begin{align*} && ax+b=xa+f(a) \\ \implies &&b=f(a) \end{align*}

    Therefore, the general solution is y=ax+f(a)y=ax+f(a). If there are initial conditions, the constant aa can be determined accurately.

  • Case 2. x+f′(y′)=0x+f^\prime (y^\prime )=0

    Organize as xx or y′y^\prime and solve the system of equations with the given differential equation.

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Below, an example is provided to understand the detailed solution process.

Example

Solve the differential equation y=xy′+(y′)2y = xy^\prime + (y^\prime )^2.

By differentiating both sides,

y′=y′+xy′′+y′′2y′  ⟹  xy′′+2y′′y′=0  ⟹  y′′[x+2y′]=0 \begin{align*} &&y^\prime =y^\prime + xy^{\prime \prime} + y^{\prime \prime}2y^\prime \\ \implies && xy^{\prime \prime} + 2y^{\prime \prime}y^\prime =0 \\ \implies && y^{\prime \prime} \left[ x+2y^\prime \right] =0 \end{align*}

  • Case 1. If it’s y′′=0y^{\prime \prime}=0

    It’s y=ax+by=ax+b and y′=ay^\prime =a. When substituted into the given differential equation, ax+b=ax+a2  ⟹  b=a2 \begin{align*} && ax+b&=ax+a^2 \\ \implies && b&=a^2 \end{align*}

    Therefore,

    y=ax+a2 y=ax+a^2

  • Case 2. If it’s x+2y′=0x+2y^\prime =0

    By organizing with respect to y′y^\prime,

    y′=112x y^\prime = 1\frac{1}{2}x

    When substituted into the given differential equation,

    y=x(−12x)+(14x2)=−12x2+14x2=−14x2 \begin{align*} y&=x\left(- \frac{1}{2}x \right) + \left( \frac{1}{4} x^2 \right) \\ &=-\frac{1}{2}x^2+\frac{1}{4}x^2 \\ &=-\frac{1}{4}x^2 \end{align*}

    Therefore,

    y=−14x2 y=-\frac{1}{4}x^2

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