1. Chapter 9 Class 12 Differential Equations
2. Serial order wise

Transcript

Ex 9.6, 7 For each of the differential equation given in Exercises 1 to 12, find the general solution : 𝑥𝑙𝑜𝑔𝑥 𝑑𝑦﷮𝑑𝑥﷯+𝑦= 2﷮𝑥﷯𝑙𝑜𝑔𝑥 Step 1 : Put in form 𝑑𝑦﷮𝑑𝑥﷯ + Py = Q xlog x 𝑑𝑦﷮𝑑𝑥﷯ + y = 2﷮𝑥﷯ log x Dividing by x log x, 𝑑𝑦﷮𝑑𝑥﷯+𝑦. 1﷮𝑥 log﷮𝑥﷯﷯ = 2﷮𝑥﷯log 𝑥 1﷮𝑥 log﷮𝑥﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯ + 1﷮𝑥 log﷮𝑥﷯﷯﷯𝑦= 2﷮𝑥﷯ Step 2 : Find P and Q Comparing (1) with 𝑑𝑦﷮𝑑𝑥﷯ + Py = Q P = 1﷮𝑥 log﷮𝑥﷯﷯ & Q = 2﷮𝑥2﷯ Step 3 : Find Integration factor, I.F I.F = e﷮ ﷮﷮𝑝 𝑑𝑥﷯﷯ I.F = e﷮ ﷮﷮ 1﷮𝑥 log﷮𝑥﷯﷯𝑑𝑥﷯﷯ Let t = log x dt = 1﷮𝑥﷯ dx dx = x dt So, I.F = e﷮ ﷮﷮ 1﷮𝑥 𝑡﷯ × 𝑥𝑑𝑡﷯﷯ I.F = e﷮ ﷮﷮ 1﷮𝑡﷯𝑑𝑡﷯﷯ I.F = e﷮ log﷮𝑡﷯﷯ I.F = 𝑡 Putting back t = log x I.F = log x Step 4 : Solution of the equation y × I.F = ﷮﷮𝑄×𝐼.𝐹. 𝑑𝑥+𝐶﷯ Putting values, y × log x = ﷮﷮ 2﷮𝑥2﷯﷯ . log x. dx + C Let I = 2 ﷮﷮ log﷮𝑥 𝑥﷮−2﷯𝑑𝑥﷯﷯ I = 2 log x. ﷮﷮ 𝑥﷮−2﷯𝑑𝑥− ﷮﷮ ﷯ 1﷮𝑥﷯ ﷮﷮ 𝑥﷮−2﷯𝑑𝑥﷯﷯﷯𝑑𝑥 ﷯ I = 2 log x . 𝑥﷮−1﷯﷮(−1)﷯ − ﷮﷮ 1﷮𝑥﷯﷯ . ( 𝑥﷮−1﷯)﷮(−1)﷯.dx ﷯ = 2 − log x. 1﷮𝑥﷯ + ﷮﷮ 1﷮ 𝑥﷮2﷯﷯.𝑑𝑥﷯﷯ = 2 − 1﷮𝑥﷯ .log x + 𝑥﷮−1﷯﷮(−1)﷯﷯ = 2 −1﷮𝑥﷯ .log x − 1﷮𝑥﷯﷯ = −2﷮𝑥﷯ (1 + log x) Putting value of I in (2) y log x = I + C y. log x = − 2﷮𝑥﷯ (1 + log x) + C y log 𝒙﷯ = −𝟐﷮𝒙﷯ (1 + log 𝒙﷯) + C Which is the general solution of the given equation. Ex 9.6, 8 For each of the differential equation given in Exercises 1 to 12, find the general solution : 1+ 𝑥﷮2﷯﷯𝑑𝑦+2𝑥𝑦 𝑑𝑥= cot﷮𝑥 𝑑𝑥 𝑥≠0﷯﷯ Step 1: Put in form 𝑑𝑦﷮𝑑𝑥﷯ + Py = Q (1 + x2)dy + 2xy.dx = cot x. dx Dividing both sides by (1 + x2) 𝑑𝑦﷮𝑑𝑥﷯ + 2𝑥𝑦﷮(1 + 𝑥2)﷯ 𝑑𝑥﷮𝑑𝑥﷯ = cot⁡(𝑥)﷮(1 + 𝑥﷮2﷯)﷯. 𝑑𝑥﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯ + 2𝑥﷮(1 + 𝑥2)﷯ y = cot﷮𝑥﷯﷮(1 + 𝑥2)﷯ Step 2 : Find P and Q Comparing (1) with 𝑑𝑦﷮𝑑𝑥﷯ + Py = Q where P = 2𝑥﷮(1 + 𝑥﷮2﷯)﷯ & Q = cot﷮𝑥﷯﷮(1 + 𝑥﷮2﷯)﷯ Step 3 : Find Integrating factor, I.F I.F. = 𝑒﷮ ﷮﷮𝑝 𝑑𝑥﷯﷯ I.F. = e﷮ ﷮﷮ 2𝑥﷮ 1 + 𝑥﷮2﷯﷯﷯ 𝑑𝑥 ﷯﷯ Let t = 1 + x2 dt = 2x dx I.F. = e﷮ ﷮﷮ 𝑑𝑡﷮𝑡﷯ ﷯﷯ = elog t = t = (1 + x2) So, I.F = 1 + x2 Step 4 : Solution of the equation y × I.F = ﷮﷮𝑄×𝐼.𝐹.𝑑𝑥+𝐶﷯ Putting values, y.(1 + x2) = ﷮﷮ cot﷮𝑥﷯﷮(1 + 𝑥2)﷯﷯ × (1+𝑥2).dx + c y.(1 + x2) = ﷮﷮ cot﷮𝑥﷯ 𝑑𝑥﷯+𝑐 y (1 + x2) = log sin﷮𝑥﷯﷯ + C Dividing by (1 + x2) y = (1 + x2)−1 log 𝐬𝐢𝐧﷮𝒙﷯﷯+𝒄 𝟏+x2﷯﷮−𝟏﷯ is the general solution of the given equation Ex 9.6, 9 For each of the differential equation find the general solution : 𝑥 𝑑𝑦﷮𝑑𝑥﷯+𝑦−𝑥+𝑥𝑦 cot﷮𝑥=0 𝑥≠0﷯﷯ Step 1: Put in form 𝑑𝑦﷮𝑑𝑥﷯ + Py = Q x 𝑑𝑦﷮𝑑𝑥﷯ + y − x + xy cot x = 0 Dividing both sides by x 𝑑𝑦﷮𝑑𝑥﷯ + 𝑦﷮𝑥﷯ − 1 + y cot x = 0 𝑑𝑦﷮𝑑𝑥﷯ + y 1﷮𝑥﷯+ cot﷮𝑥﷯﷯ − 1 = 0 𝑑𝑦﷮𝑑𝑥﷯ + 1﷮𝑥﷯+ cot﷮𝑥﷯﷯ y = 1 Step 2: Find P and Q Comparing (1) with 𝑑𝑦﷮𝑑𝑥﷯ + Py = Q P = 1﷮𝑥﷯ + cot x & Q = 1 Step 3: Find integrating factor, I.F. I.F. = e﷮ ﷮﷮𝑝 𝑑𝑥 ﷯﷯ = e﷮ ﷮﷮ 1﷮𝑥﷯ + cot﷮𝑥﷯﷯𝑑𝑥﷯﷯ = e e﷮ ﷮﷮ 1﷮𝑥﷯ 𝑑𝑥 + ﷮﷮ cot﷮𝑥﷯ 𝑑𝑥﷯﷯﷯ = 𝑒﷮ log﷮𝑥﷯ + log﷮ sin﷮𝑥﷯﷯﷯ = 𝑒﷮ log﷮𝑥 sin﷮𝑥﷯﷯﷯ = x. sin x So, I.F. = x.sin x Step 4 : Solution of the equation y × I.F. = ﷮﷮ Q×𝐼𝐹﷮𝑑𝑥﷯﷯ + C y (x sin x) = ﷮﷮𝑥. 𝑠𝑖𝑛 𝑥﷮𝑑𝑥﷯﷯ Let I = ﷮﷮𝑥. sin﷮𝑥.𝑑𝑥﷯﷯ I = x ﷮﷮ sin﷮𝑥 𝑑𝑥− ﷮﷮ 1. ﷮﷮ sin﷮𝑥 𝑑𝑥﷯﷯﷯𝑑𝑥﷯﷯﷯ = x (− cos x) − ﷮﷮1.(− cos﷮𝑥)﷯𝑑𝑥﷯ = − x. cos x + ﷮﷮ cos﷮𝑥 𝑑𝑥﷯﷯ = − x cos x + sin x Putting value of I in (2), y.x. sin x = −x.cos x + sin x + C Divide by x sin x y = 𝑥 cos﷮𝑥﷯﷮𝑥 sin﷮𝑥﷯﷯ + sin﷮𝑥﷯﷮𝑥 sin﷮𝑥﷯﷯ + 𝑐﷮𝑥 sin﷮𝑥﷯﷯ y = −cot x + 𝟏﷮𝒙﷯ + 𝒄﷮𝒙 𝒔𝒊𝒏﷮𝒙﷯﷯ Which is the general solution of the given differential equation