Question

Substitute$(7.22)$into$(7.21)$to obtain the equation for$v^{\text{'}}\left(x\right)$. Show that this equation is separable.

Short answer

Therefore, the equation for$v^{\text{'}}\left(x\right)$is$\frac{\left[\begin{array}{c}-u\left(x\right)v^{\text{'}\text{'}}\left(x\right)-u^{\text{'}\text{'}}\left(x\right)v\left(x\right)-\\ \left(f\left(x\right)u^{\text{'}}\left(x\right)v\left(x\right)+g\left(x\right)u\left(x\right)v\left(x\right)\right)\end{array}\right]}{\left[\left(2u^{\text{'}}\left(x\right)+f\left(x\right)u\left(x\right)\right)v^{\text{'}}\left(x\right)\right]}$and it is separable.

Step-by-step solution

Given information

Consider equation (7.22)

$y=u\left(x\right)v\left(x\right)$

Separable equation

A separable differential equation is any equation that can be written in the form $\mathbf{y}\text{'}=\mathbf{f}\left(\mathbf{x}\right)\mathbf{g}\left(\mathbf{y}\right).$

Solve for  v'(x)

Consider equation (7.21)$y^{\text{'}\text{'}}+f\left(x\right)y^{\text{'}}+g\left(x\right)y=0\dots \text{(1)}$

So,

$$\begin{gathered} y=u\left(x\right)v\left(x\right) \\ {y}^{\text{'}}=u\left(x\right)v^{\text{'}}\left(x\right)+u^{\text{'}}\left(x\right)v\left(x\right) \end{gathered}$$

And,

$y^{\text{'}\text{'}}=2u^{\text{'}}\left(x\right)v^{\text{'}}\left(x\right)+u\left(x\right)v^{\text{'}\text{'}}\left(x\right)+u^{\text{'}\text{'}}\left(x\right)v\left(x\right)$

Substitute the value of $y,y^{\text{'}}$and $y^{\text{'}\text{'}}$in equation (1).

$\left.\left(2u^{\text{'}}\left(x\right)v^{\text{'}}\left(x\right)+u\left(x\right)v^{\text{'}\text{'}}\left(x\right)+u^{\text{'}\text{'}}\left(x\right)v\left(x\right)\right)+f\left(x\right)u\left(x\right)v^{\text{'}}\left(x\right)+u^{\text{'}}\left(x\right)v\left(x\right)+\right]g\left(x\right)u\left(x\right)v\left(x\right)$

$$\begin{gathered} \left[\begin{array}{c}\left(2x{u}^{\text{'}}\left(x\right)v^{\text{'}}\left(x\right)+f\left(x\right)u\left(x\right)v^{\text{'}}\left(x\right)\right)+u\left(x\right)v^{\text{'}\text{'}}\left(x\right)+u^{\text{'}\text{'}}\left(x\right)v\left(x\right)+\\ \left(f\left(x\right)u^{\text{'}}\left(x\right)v\left(x\right)+g\left(x\right)u\left(x\right)v\left(x\right)\right)\end{array}\right]=0 \\ \left[\left(2u^{\text{'}}\left(x\right)+f\left(x\right)u\left(x\right)\right)v^{\text{'}}\left(x\right)\right]=\left[\begin{array}{c}-u\left(x\right)v^{\text{'}\text{'}}\left(x\right)-u^{\text{'}\text{'}}\left(x\right)v\left(x\right)-\\ \left(f\left(x\right)u^{\text{'}}\left(x\right)v\left(x\right)+g\left(x\right)u\left(x\right)v\left(x\right)\right)\end{array}\right] \end{gathered}$$

Solve further,

$v^{\text{'}}\left(x\right)=\frac{\left[\begin{array}{c}-u\left(x\right)v^{\text{'}\text{'}}\left(x\right)-u^{\text{'}\text{'}}\left(x\right)v\left(x\right)-\\ \left(f\left(x\right)u^{\text{'}}\left(x\right)v\left(x\right)+g\left(x\right)u\left(x\right)v\left(x\right)\right)\end{array}\right]}{\left[\left(2u^{\text{'}}\left(x\right)+f\left(x\right)u\left(x\right)\right)v^{\text{'}}\left(x\right)\right]}$

Therefore, the equation for $v^{\text{'}}\left(x\right)$is $\frac{\left[\begin{array}{c}-u\left(x\right)v^{\text{'}\text{'}}\left(x\right)-u^{\text{'}\text{'}}\left(x\right)v\left(x\right)-\\ \left(f\left(x\right)u^{\text{'}}\left(x\right)v\left(x\right)+g\left(x\right)u\left(x\right)v\left(x\right)\right)\end{array}\right]}{\left[\left(2u^{\text{'}}\left(x\right)+f\left(x\right)u\left(x\right)\right)v^{\text{'}}\left(x\right)\right]}$and the proof that it is separable is stated above.