2.II.6 B2 . \mathrm{II} . 6 \mathrm{~B} \quad

Differential Equations | Part IA, 2001

The function y(x)y(x) satisfies the linear equation

y(x)+p(x)y(x)+q(x)y(x)=0y^{\prime \prime}(x)+p(x) y^{\prime}(x)+q(x) y(x)=0

The Wronskian, W(x)W(x), of two independent solutions denoted y1(x)y_{1}(x) and y2(x)y_{2}(x) is defined to be

W(x)=y1y2y1y2.W(x)=\left|\begin{array}{cc} y_{1} & y_{2} \\ y_{1}^{\prime} & y_{2}^{\prime} \end{array}\right| .

Let y1(x)y_{1}(x) be given. In this case, show that the expression for W(x)W(x) can be interpreted as a first-order inhomogeneous differential equation for y2(x)y_{2}(x). Hence, by explicit derivation, show that y2(x)y_{2}(x) may be expressed as

y2(x)=y1(x)x0xW(t)y1(t)2dt,y_{2}(x)=y_{1}(x) \int_{x_{0}}^{x} \frac{W(t)}{y_{1}(t)^{2}} d t,

where the rôle of x0x_{0} should be briefly elucidated.

Show that W(x)W(x) satisfies

dW(x)dx+p(x)W(x)=0.\frac{d W(x)}{d x}+p(x) W(x)=0 .

Verify that y1(x)=1xy_{1}(x)=1-x is a solution of

xy(x)(1x2)y(x)(1+x)y(x)=0.x y^{\prime \prime}(x)-\left(1-x^{2}\right) y^{\prime}(x)-(1+x) y(x)=0 .

Hence, using ()(*) with x0=0x_{0}=0 and expanding the integrand in powers of tt to order t3t^{3}, find the first three non-zero terms in the power series expansion for a solution, y2(x)y_{2}(x), of ( \dagger ) that is independent of y1(x)y_{1}(x) and satisfies y2(0)=0,y2(0)=1y_{2}(0)=0, y_{2}{ }^{\prime \prime}(0)=1.

Typos? Please submit corrections to this page on GitHub.