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Prove the following.
If y = e^(a cos-1 x), where -1 ≤ x ≤ 1, then show that: (1 - x²) y₂ - xy₁ - a²y = 0
If $y = e^{a \cos^{-1} x}$, where $-1 \leq x \leq 1$, then show that:
$(1 - x^2) y_2 - xy_1 - a^2y = 0$
Given: $y = e^{a \cos^{-1} x}$, where $-1 \leq x \leq 1$
To show: $(1 - x^2) y_2 - xy_1 - a^2y = 0$
Answer
Answer
$y = e^{a\cos^{-1}x}$
$y_1 = e^{a\cos^{-1}x}\cdot\left(\frac{-a}{\sqrt{1-x^2}}\right) = \frac{-ay}{\sqrt{1-x^2}}$
$\sqrt{1-x^2}\,y_1 = -ay$; squaring: $(1-x^2)y_1^2 = a^2y^2$.
Differentiating: $-2xy_1^2 + 2(1-x^2)y_1y_2 = 2a^2yy_1$.
Dividing by $2y_1$: $(1-x^2)y_2 - xy_1 = a^2y$.
Hence $(1-x^2)y_2 - xy_1 - a^2y = 0$.
- $y_1 = -\frac{ay}{\sqrt{1-x^2}}$, by the chain rule.
- $(1-x^2)y_1^2 = a^2y^2$, by squaring.
- $-2xy_1^2 + 2(1-x^2)y_1y_2 = 2a^2yy_1$, differentiating with respect to $x$.
- Divide by $2y_1$: $-xy_1 + (1-x^2)y_2 = a^2y$.
- Hence $(1-x^2)y_2 - xy_1 - a^2y = 0$.