Ex 9.6, 14

Transcript

Ex 9.6, 14 For each of the differential equations given in Exercises 13 to 15 , find a particular solution satisfy the given condition : 1+ 𝑥﷮2﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯+2𝑥𝑦= 1﷮1+ 𝑥﷮2﷯﷯ ;𝑦=0 when 𝑥=1 (1 + x2) 𝑑𝑦﷮𝑑𝑥﷯ + 2xy = 1﷮1 + 𝑥2﷯ Divide both sides by (1+𝑥2) 𝑑𝑦﷮𝑑𝑥﷯ + 2𝑥𝑦﷮1 + 𝑥﷮2﷯﷯ = 1﷮ 1 + 𝑥2﷯.(1 + 𝑥2)﷯ 𝑑𝑦﷮𝑑𝑥﷯ + 2𝑥𝑦﷮1 + 𝑥﷮2﷯﷯﷯y = 1﷮ 1 + 𝑥2﷯﷯ Comparing with 𝑑𝑦﷮𝑑𝑥﷯ + Py = Q P = 2𝑥﷮1 + 𝑥﷮2﷯﷯ & Q = 1﷮ 1 + 𝑥2﷯2﷯ Find Integrating factor IF = 𝑒﷮ ﷮﷮𝑝 𝑑𝑥﷯﷯ IF = 𝑒﷮ ﷮﷮ 2𝑥﷮1 + 𝑥﷮2﷯﷯ 𝑑𝑥﷯﷯ Let 1+ 𝑥﷮2﷯ = t Diff . w.r.t. x 2x = 𝑑﷮𝑑𝑥﷯ t dx = 𝑑𝑡﷮2𝑥﷯ IF = e﷮ ﷮﷮ 2𝑥﷮𝑡﷯﷯ 𝑑𝑡﷮2𝑥﷯﷯ IF = e﷮ ﷮﷮ 𝑑𝑡﷮𝑡﷯﷯﷯ IF = e﷮𝑙𝑜𝑔 𝑡﷯﷯ IF = t IF = 1 + x2 Step 4 : Solution of the deferential equation y × I.F = ﷮﷮𝑄×𝐼.𝐹 𝑑𝑥﷯ Putting values y × (1 + x2) = ﷮﷮ 1﷮ 1 + 𝑥﷮2﷯﷯﷮2﷯﷯﷯(1 + x2).dx y × (1 + x2) = ﷮﷮ 1﷮ 1 + 𝑥﷮2﷯﷯﷯﷯dx y (1 + x2) = tan﷮−1﷯﷮ 𝑥+𝑐﷯ Given that y = 0 when x = 1 Putting y = 0 and x = 1 in (1) y (1 + x2) = tan−1 x + c 0(1 + 12) = tan−1 (1)+ c 0 = 𝜋﷮4﷯ + C C = − 𝜋﷮4﷯ Putting value of C in (2) y (1 + x2) = tan-1 x + c y (1 + x2) = tan-1 x − 𝝅﷮𝟒﷯