Question

यदि y= e^{a cos^ {-1 x}}, -1 ≤ x≤ 1 तो दर्शाइए कि
(1-x^{2}) d^{2}y/dx^{2} -x dy/dx - a^{2}y =0

Answer 1

(1 - x²)(d²y/dx²)   + xdy/dx   - a²y = 0

Step-by-step explanation:

$y= e^{a cos^ {-1}x}$

$(1-x^{2}) d^{2}y/dx^{2} + x dy/dx - a^{2}y =0$

y = eᵇ

b = aCos⁻¹x

dy/dx = eᵇ (db/dx)

b = aCos⁻¹x

db/dx = -a/√(1 - x²)

dy/dx = eᵇ (-a/√(1 - x²))

√(1 - x²) dy/dx = - ay

√(1 - x²)(d²y/dx ²)   + ((-1/2√(1 - x²)(-2x)))dy/dx   = -ady/dx

√(1 - x²)(d²y/dx²)   + x/√(1 - x²)dy/dx   = -ady/dx

(1 - x²)(d²y/dx²)   + xdy/dx   = -a√(1 - x²)dy/dx

( ∵  √(1 - x²) dy/dx = - ay  => dy/dx = -ay/√(1 - x²))

(1 - x²)(d²y/dx²)   + xdy/dx   = -a√(1 - x²)(-ay/√(1 - x²))

(1 - x²)(d²y/dx²)   + xdy/dx   = a²y

(1 - x²)(d²y/dx²)   + xdy/dx   - a²y = 0

और अधिक जानें :

(x + 3)^{2} .(x + 4)^{3} .(x + 5)^{4} प्रदत्त फलनों का x के सापेक्ष अवकलन कीजिए

brainly.in/question/15287089

f(x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) द्वारा प्रदत्त फलन का अवकलज ज्ञात कीजिए और इस प्रकार f'(1) ज्ञात कीजिए।

brainly.in/question/15287093