Misc 7 - Chapter 9 Class 12 Differential Equations

Transcript

Misc 7 Show that the general solution of the differential equation 𝑑𝑦﷮𝑑𝑥﷯+ 𝑦﷮2﷯+𝑦+1﷮ 𝑥﷮2﷯+𝑥+1﷯=0 is given by 𝑥+𝑦+1﷯=A 1−𝑥−𝑦−2𝑥𝑦﷯, where A is parameter. 𝑑𝑦﷮𝑑𝑥﷯+ 𝑦﷮2﷯ + 𝑦 + 1﷮ 𝑥﷮2﷯ + 𝑥 + 1﷯ = 0 𝑑𝑦﷮𝑑𝑥﷯= −( 𝑦﷮2﷯ + 𝑦 + 1)﷮ 𝑥﷮2﷯ + 𝑥 + 1﷯ 𝑑𝑦﷮ 𝑦﷮2﷯ + 𝑦 + 1﷯= −𝑑𝑥﷮ 𝑥﷮2﷯ + 𝑥 + 1﷯ 𝑑𝑦﷮𝑦 +2 1﷮2﷯﷯𝑦 + 1﷮2﷯﷯﷮2﷯− 1﷮2﷯﷯﷮2﷯+ 1﷯= −𝑑𝑥﷮𝑦 +2 1﷮2﷯﷯𝑦 + 1﷮2﷯﷯﷮2﷯− 1﷮2﷯﷯﷮2﷯+ 1﷯ 𝑑𝑦﷮ 𝑦 + 1﷮2﷯﷯﷮2﷯+ 3﷮4﷯﷯= −𝑑𝑥﷮ 𝑥 + 1﷮2﷯﷯﷮2﷯+ 3﷮4﷯﷯ 𝑑𝑦﷮ 𝑦 + 1﷮2﷯﷯﷮2﷯+ ﷮3﷯﷮2﷯﷯﷮2﷯﷯= −𝑑𝑥﷮ 𝑥 + 1﷮2﷯﷯﷮2﷯+ ﷮3﷯﷮2﷯﷯﷮2﷯﷯ Integrating both sides ﷮﷮ 𝑑𝑦﷮ 𝑦 + 1﷮2﷯﷯﷮2﷯ + ﷮3﷯﷮2﷯﷯﷮2﷯ ﷯﷯ = − ﷮﷮ 𝑑𝑥﷮ 𝑥 + 1﷮2﷯﷯﷮2﷯ + ﷮3﷯﷮2﷯﷯﷮2﷯ ﷯﷯ 2﷮ ﷮3﷯﷯ tan−1 𝑦 + 1﷮2﷯﷮ ﷮3﷯﷮2﷯﷯﷯ = −2﷮ ﷮3﷯﷯ tan−1 𝑥 + 1﷮2﷯﷮ ﷮3﷯﷮2﷯﷯﷯ 2﷮ ﷮3﷯﷯ tan−1 2𝑦 + 1﷮ ﷮3﷯﷯﷯ + tan−1 2𝑥 + 1﷮ ﷮3﷯﷯﷯﷯ = C 2﷮ ﷮3﷯﷯ tan−1 2𝑦 + 1﷮ ﷮3﷯﷯+ 2𝑥 + 1﷮ ﷮3﷯﷯﷮1− 2𝑦 − 1﷮ ﷮3﷯﷯ × 2𝑥 + 1﷮ ﷮3﷯﷯ ﷯﷯=𝐶 2𝑦 + 1 + 2𝑥 + 1﷮ ﷮3﷯﷯﷮1− (2𝑦 + 1)(2𝑥 + 1)﷮ ﷮3﷯﷯﷯ = tan ﷮3﷯﷮2﷯𝐶﷯ 2﷮ ﷮3﷯﷯ tan−1 2𝑦 + 1﷮2﷯﷮ ﷮3﷯﷮2﷯﷯﷯ = − 2﷮ ﷮3﷯﷯ tan−1 2𝑥 + 1﷮2﷯﷮ ﷮3﷯﷮2﷯﷯﷯ = C 2﷮ ﷮3﷯﷯ tan−1 2𝑦 + 1﷮ ﷮3﷯﷯﷯ + tan−1 2𝑥 + 1﷮ ﷮3﷯﷯﷯﷯ = C 2﷮ ﷮3﷯﷯ tan−1 2𝑦 + 1﷮ ﷮3﷯﷯ + 2𝑥 + 1﷮ ﷮3﷯﷯﷮1 − 2𝑦 − 1﷮ ﷮3﷯﷯ × 2𝑥 + 1﷮ ﷮3﷯﷯ ﷯﷯=𝐶 2𝑦 + 1 + 2𝑥 + 1﷮ ﷮3﷯﷯﷮1 − (2𝑦 + 1)(2𝑥 + 1)﷮3﷯﷯ = tan ﷮3﷯﷮2﷯𝐶﷯ 2𝑦 + 1 + 2𝑥 + 1﷮ ﷮3﷯﷯﷮ 3 − (2𝑦 + 1)(2𝑥 + 1)﷮3﷯﷯ = C1 ﷮3﷯(2𝑦 + 2𝑥 + 2)﷮3 − 4𝑥𝑦 + 2𝑦 + 2𝑥 + 1﷯﷯ = C 2 ﷮3﷯ (x + y + 1) = C1 3−4𝑥𝑦−2𝑥−2𝑦−1﷯ 2 ﷮3﷯ (x + y + 1) = C1 2−4𝑥𝑦−2𝑥−2𝑦﷯ ﷮𝟑﷯ (x + y + 1) = C1 ( 1 − x − y − 2xy) is the required general solution