Misc 13 - Chapter 9 Class 12 Differential Equations (Term 2)

Transcript

Misc 13 Find a particular solution of the differential equation 𝑑𝑦﷮𝑑𝑥﷯+𝑦 cot﷮𝑥=4𝑥 𝑐𝑜𝑠𝑒𝑐 𝑥 𝑥≠0﷯ ,﷯ given that 𝑦=0 when 𝑥= 𝜋﷮2﷯ Given 𝑑𝑦﷮𝑑𝑥﷯+𝑦 cot﷮𝑥=4𝑥 𝑐𝑜𝑠𝑒𝑐 𝑥 ﷯ This of the form 𝑑𝑦﷮𝑑𝑥﷯+𝑃𝑦=𝑄 where P = cot x & Q = 4x cosec x IF = 𝑒﷮ ﷮﷮𝑃𝑑𝑥﷯﷯ IF = 𝑒﷮ ﷮﷮ cot﷮𝑥 𝑑𝑥﷯﷯﷯ IF = 𝑒﷮log⁡( sin﷮𝑥)﷯﷯ IF = sin x Solution is y (IF) = ﷮﷮ 𝑄×𝐼.𝐹﷯𝑑𝑥+𝑐 ﷯ y sin x = ﷮﷮4𝑥 𝑐𝑜𝑠𝑒𝑐 𝑥 sin﷮𝑥﷯ 𝑑𝑥+𝑐 ﷯ y sin x = ﷮﷮4𝑥 1﷮ sin﷮𝑥﷯﷯ sin﷮𝑥﷯𝑑𝑥+𝑐 ﷯ y sin x = ﷮﷮4𝑥 𝑑𝑥+𝑐 ﷯ y sin x = 4 𝑥﷮2﷯﷮2﷯+𝑐 y sin x = 2x2 + C Given that 𝑦=0 when 𝑥= 𝜋﷮2﷯ Put x = 𝜋﷮2﷯ & y = 0 in (1) 0 × sin 𝜋﷮2﷯ = 2 𝜋﷮2﷯﷯﷮2﷯+𝐶 0 = 2 𝜋﷮4﷯﷮2﷯﷯ + C 0 = 𝜋﷮2﷯﷮2﷯ + C C = −𝜋﷮2﷯﷮2﷯ Putting value of C in (1) y sin x = 2x2 + c y sin x = 2 𝒙﷮𝟐﷯ − 𝝅﷮𝟐﷯﷮𝟐﷯