Question
Solve: .
Answer :
Word Count : 448
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Skip in 10
We are asked to solve the differential equation: [ x^2 y'' - 2x y' - 4y = x^2 + 2 \ln x ] Step 1: Solve the corresponding homogeneous equation [ x^2 y'' - 2x y' - 4y = 0 ] This is a Cauchy-Euler (equidimensional) equation. We assume a solution of the form (y = x^m). Then [ y' = m x^{m-1}, \quad y'' = m(m-1)x^{m-2} ] Substitute into the homogeneous equation: [ x^2[m(m-1)x^{m-2}] - 2x[m x^{m-1}] - 4 x^m = 0 ] Simplify each term: [ m(m-1)x^m - 2 m x^m - 4 x^m = 0 ] Combine terms: [ [m(m-1) - 2m - 4] x^m = 0 ] [ [m^2 - m - 2m - 4] x^m = 0 ] [ [m^2 - 3m - 4] x^m = 0 ] Characteristic equation: [ m^2 - 3m - 4 = 0 ] Solve quadratic: [ m = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm 5}{2} ] [ m_1
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Related Questions
हल कीजिए : .
Suppose that a thermometer having a reading of inside a house is placed outside where the air temperature is
. Two minutes later it is found that the thermometer reading is
. Find the temperature reading T(t) of the thermometer at any time t.
मान लीजिए एक थर्मामीटर जिसके घर के अंदर रीडिंग है, उसे बाहर रखा जाता है, जहाँ वायु तापमान
है। दो मिनट के बाद थर्मामीटर की रीडिंग
पाई जाती है। किसी भी समय t पर थर्मामीटर के तापमान की रीडिंग T(t) ज्ञात कीजिए।
Solve: .
हल कीजिए :
(2)
in
is a linear homogeneous equation
हल कीजिए :