# Ex 9.5, 4 - Chapter 9 Class 12 Differential Equations

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 9.5, 4show that the given differential equation is homogeneous and solve each of them.( ^2 ^2 ) +2 =0 Step 1: Find / ( ^2 ^2 ) +2 =0 2xy dy = ( ^2 ^2 ) dx 2xy dy = ( ^2 ^2 ) dx / = ( ^2 ^2)/2 Step 2: Putting F(x, y) = / and finding F( x, y) F(x, y) = ( ^2 ^2)/2 F( x, y) = (( )^2 ( )^2)/(2 . )= ( ^2 ^2 ^2 ^2)/( ^2.2 )= ( ^2 ( ^2 ^2))/( ^2.2 ) = ( ^2 ^2)/2 = F(x, y) F( x, y) = F(x, y) = F(x, y) Hence, F(x, y) is a homogenous Function of with degree zero So, / is a homogenous differential equation. Step 3. Solving / by putting y = vx Put y = vx. differentiating w.r.t.x / = x / + / / = / + v Putting value of / and y = vx in (1) / = ( ^2 ^2)/2 ( )/ + = ( ( ) ^2 ^2)/(2 ( )) ( )/ + = ( ^2 ^2 ^2)/(2 ^2 ) ( )/ = ( ^2 ^2 ^2)/(2 ^2 ) v x / = ( ^2 ^2 ^2 2 ^2 ^2)/(2 ^2 ) x / = ( ^2 ^2 ^2)/(2 ^2 ) / = 1/ (( ^2 ( ^2 + 1))/(2 ^2 )) / = 1/ (( ^2 + 1)/2 ) (2 )/( ^2 + 1) = ( )/ Integrating both sides 1 2 /( ^2 + 1) = 1 ( )/ 1 2 /( ^2 + 1) = log | |+ Putting t = v2 + 1 diff.w.r.t v. / (v2 + 1) = / 2v = / dv = /2 Now, From (2) 1 2 / /2 = log| |+ ( from (2) ) 1 / = log| |+ log| | = log| |+ Putting t = v2 + 1 log|"v2 + 1" | = log|"x" |+ log |"v2 + 1" | + log|"x" |= log |"x(v2 + 1)" | = C Putting v = / log |[( / )^2+1] |= log|( ^2 + ^2)/ ^2 |= Putting = log c log |( ^2 + ^2)/ | = log c ^2+ ^2 = cx ^ + ^ = is the general solution of the given differential equation.

Chapter 9 Class 12 Differential Equations

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.