Answer to Question #264887 in Differential Equations for tina

Question

Let 𝑦1 and 𝑦2 be linearly independent solutions of the differential equation 𝑦 ′′ + 𝑝(𝑥)𝑦 ′ + 𝑞(𝑥)𝑦 = 0, where functions 𝑝 and 𝑞 are continuous on some interval 𝐼. (i) Prove that 𝑊(𝑦1, 𝑦2 )(𝑥) = 𝐶𝑒 − ∫ 𝑝(𝑥)𝑑𝑥 , where 𝑊 is the Wronskian and 𝐶 ∈ ℝ is an arbitrary constant.

Answer 1

$y''+p(x)y''+q(x)y=0$

Since y₁(x) and y₂(x) are linear independent solutions of the differential equation above, we get:

$y_1''+p(x)y_1''+q(x)y_1=0$ .........(1)

$y_2''+p(x)y_2''+q(x)y_2=0$ ..........(2)

Now from (1) we get,

$y_1''=-(p(x)y_1'+y_1)$ .......(3)

From (2) we get,

$y_2''=-(p(x)y_2'+y_2)$ .........(4)

Now we know that the Wronskian

$W(y_1,y_2)=\begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix}=y_1y_2'-y_2y_1'$

$W=y_1y_2'-y_2y_1'$

$W'=(y_1y_2'-y_2y_1')'$

$W'=(y_1y_2')'-(y_2y_1')'$

$W'=(y_1 \cdot y_2''+y_2'\cdot y_1')-(y_2\cdot y_1''+y_1'\cdot y_2')$

Since $(uv)'=u\cdot v'+v\cdot u'$

$W'=y_1 \cdot y_2''+y_2'\cdot y_1'-y_2\cdot y_1''-y_1'\cdot y_2'$

$W'=y_1\cdot[-p(x)y_2'+y_2)]-y_2\cdot[-(p(x)y_1'+y_1)]$

Values from (1)and (2)

$W'=-p(x)y_1y_2'-y_1y_2+p(x)y_2y_1'+y_1y_2$

$W'=-p(x)y_1y_2'+p(x)y_2y_1'$

$W'=-p(x)[y_1y_2'-y_2y_1']$

$W'=-p(x)W$

This is an equation of the form z'=-pz whose solution is in the form $z=ce^{-\int pdx}$

From $W'=-pW$ we get

$W(y_1,y_2)=ce^{-\int Pdx}$