Examples

Chapter 9 Class 12 Differential Equations
Serial order wise

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### Transcript

Example 2 Verify that the function π¦=π^(β3π₯) is a solution of the differential equation (π^2 π¦)/(ππ₯^2 )+ππ¦/ππ₯β6π¦=0 π¦=π^(β3π₯) ππ/ππ=π(π^(β3π₯) )/ππ₯ ππ¦/ππ₯=γβ3 πγ^(β3π₯) (π^π π)/(ππ^π )=π/ππ₯ (ππ¦/ππ₯) =π(γβ3 πγ^(β3π₯) )/ππ₯ =β3 π(π^(β3π₯) )/ππ₯ =β3 Γ (γβ3 πγ^(β3π₯) ) = γ9 πγ^(β3π₯) Now, we have to verify (π^2 π¦)/(ππ₯^2 )+ππ¦/ππ₯β6π¦=0 Solving L.H.S (π^2 π¦)/(ππ₯^2 )+ππ¦/ππ₯β6π¦ Putting values = γ9 πγ^(β3π₯)+(β3π^(β3π₯) )β6(π^(β3π₯) ) =γ9 πγ^(β3π₯)β3π^(β3π₯)β6π^(β3π₯) =γ9 πγ^(β3π₯)β9π^(β3π₯) =0 = R.H.S β΄ Hence Verified