📂Odinary Differential Equations

Methods for Finding the Second Solution of Second Order Differential Equations

Description¹

$$ \begin{equation} y^{\prime \prime }+p(t)y^{\prime} + q(t)y=0 \end{equation} $$

Given the differential equation above, assume we know one solution $y_{1}$. Let’s assume the general solution is $y(t)=\nu (t) y_{1}(t)$. If we calculate the 1st and 2nd derivatives of $y$, we get the following.

$$ \begin{align*} y^{\prime} &= \nu^{\prime} y_{1} + \nu y_{1}^{\prime} \\ y^{\prime \prime} &= \nu ^{\prime \prime}y_{1} + \nu^{\prime} y_{1}^{\prime} + \nu^ \prime y_{1}^{\prime} + \nu y_{1}^{\prime \prime} \\ &= \nu ^{\prime \prime}y_{1} + 2\nu^{\prime} y_{1}^{\prime} + \nu y_{1}^{\prime \prime} \end{align*} $$

By substituting $y^{\prime}$ and $y^{\prime \prime}$ into $(1)$,

$$ \nu^{\prime \prime}y_{1} + 2\nu^{\prime} y_{1}^{\prime} + \nu y_{1} ^{\prime \prime} + p \left( \nu^{\prime} y_{1} + \nu y_{1}^{\prime} \right) + q\nu y_{1}=0 $$

After arranging for $\nu$,

$$ \begin{equation} \nu \left( y_{1} ^{\prime \prime} + py_{1}^{\prime} + qy_{1} \right) + \nu^{\prime} \left( 2y_{1}^{\prime} + py_{1} \right) + \nu ^{ \prime \prime} y_{1}=0 \end{equation} $$

Since $y_{1}$ is a solution of $(1)$, it follows that $y_{1}^{\prime \prime} + py_{1}^ \prime + qy=0$. Therefore, the first term of $(2)$ is $0$, and upon rearranging, $$ \nu^{\prime} \left( 2y_{1}^{\prime} + py_{1} \right) + \nu ^{ \prime \prime}y_{1}=0 $$

To reduce the coefficients of the differential equation, let’s substitute with $\nu^{\prime} \equiv w$. Then, the equation becomes a first-order differential equation with reduced coefficients.

$$ w \left( 2y_{1}^{\prime}+ py_{1} \right) + w^{\prime} y_{1}=0 $$

By solving the newly obtained differential equation for $w$ using techniques like separation of variables, we can find the second solution and the general solution. Let’s see this through an example.

Example

Find the second solution and the general solution when $2t^2 y^{\prime \prime} + 3ty^{\prime} –y=0$, $t>0$, $y_{1}=t^{-1}$.

If we assume $y=\nu t^{-1}$,

$$ y^{\prime} = \nu^{\prime} t^{-1} - \nu y^{-2} \\ y^{\prime \prime } = \nu ^{\prime \prime} t^{-1} - 2v^{\prime} t^{-2}+ 2\nu t^{-3} $$

By substituting into the given differential equation,

$$ 2t^2 \left( \nu ^{\prime \prime} t^{-1} + 2\nu t^{-3} - 2\nu^{\prime} t^{-2} \right) +3t \left( \nu^{\prime} t^{-1} - \nu t^{-2} \right) -\nu t^{-1} = 0 $$

After arranging for $\nu$, since the term $\nu$ is $0$,

$$ 2t\nu^{\prime \prime} -\nu^{\prime} =0 $$

Once we substitute with $\nu^{\prime} \equiv w$,

$$ \begin{align*} && 2tw^{\prime} –w&=0 \\ \implies && 2t w^{\prime} &=0 \\ \implies && \dfrac{1}{w}dw &= \dfrac{1}{2t}dt \\ \implies && \ln w &= \dfrac{1}{2} \ln t + C = \ln t ^{1/2} +C \\ \implies && w &= Ce^{\ln t^{1/2}} = Ct^{1/2} \\ \implies && \nu^{\prime} &= w=Ct^{1/2} \\ \implies && \nu &= \frac{2}{3}Ct^{\frac{3}{2}} + k \end{align*} $$

Therefore, $y=\nu y_{1}=\nu t^{-1}$,

$$ \begin{align*} && y &=\left( \frac{2}{3}Ct^{\frac{3}{2}} + k \right) t^{-1} \\ \implies && y&=\frac{2}{3}Ct^{\frac{1}{2}} + kt^{-1} \end{align*} $$

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