Verifying Solutions to a Differential Equation

1. Show that \(y=f(x)=x^{4}-5x+1\) is a solution of the differential equation \(y^{\prime \prime} -12x^{2}=0\).
2. Show that \(y=g(x)=x^{4}+C_{1}x+C_{2},\) where \(C_{1}\) and \(C_{2}\) are constants, is a solution* of the differential equation \(y^{\prime \prime} -12x^{2}=0\).
3. Show that \(x^{2}-x^{3}y+3y^{4}=C,\) where \(C\) is a constant, is a solution* of the differential equation \(\dfrac{dy}{dx}=\dfrac{3x^{2}y-2x}{12y^{3}-x^{3}}.\)

Solution (a) For \(y=f(x)=x^{4}-5x+1,\) we have \[ \begin{equation*} y^{\prime} =4x^{3}-5\qquad y^{\prime \prime} =12x^{2} \end{equation*} \]

The function \(f\) satisfies the differential equation \(y^{\prime \prime} -12x^{2}=0\) and so \(f\) is a solution.

(b) For \(y=g(x)=x^{4}+C_{1}x+C_{2}\), we have \[ \begin{equation*} y^{\prime} =4x^{3}+C_{1}\qquad y^{\prime \prime} =12x^{2} \end{equation*} \]

The function \(g\) satisfies the differential equation \(y^{\prime \prime} -12x^{2}=0\), so \(g\) is a solution.

(c) Differentiate \(x^{2}-x^{3}y+3y^{4}=C\) implicitly with respect to \(x\) to find \(\dfrac{dy}{dx}\). \[ \begin{eqnarray*} 2x-x^{3}\dfrac{dy}{dx}-3x^{2}y+12y^{3}\dfrac{dy}{dx} &=&0 \\ ( 12y^{3}-x^{3}) \dfrac{dy}{dx} &=&3x^{2}y-2x \\[3pt] \dfrac{dy}{dx} &=&\dfrac{3x^{2}y-2x}{12y^{3}-x^{3}} \end{eqnarray*} \]

The function \(y=f( x) \) defined by the equation \(x^{2}-x^{3}y+3y^{4}=C\) satisfies the first-order differential equation and so is a solution.