Example 19 - Chapter 9 Class 12 Differential Equations
Last updated at April 16, 2024 by Teachoo
Gen and Particular Solution
Gen and Particular Solution
Last updated at April 16, 2024 by Teachoo
Example 19 Verify that the function π¦=π1 π^ππ₯ cosβ‘γππ₯+π2 π^ππ₯ sinβ‘ππ₯ γ , π€βπππ π1 , π2 are arbitrary constants is a solution of the differential equation (π^2 π¦)/(ππ₯^2 )β2π ππ¦/ππ₯+(π^2+π^2 )π¦=0 π¦=π1 π^ππ₯ cosβ‘γππ₯+π2 π^ππ₯ sinβ‘γππ₯, γ γ π =π^ππ (ππ πππ ππ+ ππ πππβ‘ππ) Differentiating w.r.t. x π¦^β²=(π^ππ₯ )^β² (π_1 cosβ‘ππ₯+π_2 sinβ‘ππ₯ )+π^ππ₯ (π_1 cosβ‘ππ₯+π_2 sinβ‘ππ₯ )^β² π¦^β²=ππ^ππ₯ (π_1 cosβ‘ππ₯+π_2 sinβ‘ππ₯ )+π^ππ₯ (γβπγ_1 π sinβ‘ππ₯+π_2 π cosβ‘ππ₯ ) π¦^β²=ππ^ππ₯ (π_1 cosβ‘ππ₯+π_2 sinβ‘ππ₯ )+ππ^ππ₯ (γβπγ_1 sinβ‘ππ₯+π_2 cosβ‘ππ₯ ) Putting π¦ =π^ππ₯ (π1 πππ ππ₯+ π2 sinβ‘ππ₯) π¦^β²=ππ¦+ππ^ππ₯ (γβπγ_1 sinβ‘ππ₯+π_2 cosβ‘ππ₯ ) π^β²βππ=ππ^ππ (γβπγ_π πππβ‘ππ+π_π πππβ‘ππ ) Differentiating again w.r.t x π¦^β²β²βππ¦^β²=(ππ^ππ₯ )^β² (γβπγ_1 sinβ‘ππ₯+π_2 cosβ‘ππ₯ )+ππ^ππ₯ (γβπγ_1 sinβ‘ππ₯+π_2 cosβ‘ππ₯ )^β² π¦^β²β²βππ¦^β²=πππ^ππ₯ (γβπγ_1 sinβ‘ππ₯+π_2 cosβ‘ππ₯ )+ππ^ππ₯ (γβπγ_1 π cosβ‘ππ₯βπ_2 π sinβ‘ππ₯ ) π¦^β²β²βππ¦^β²=πππ^ππ₯ (γβπγ_1 sinβ‘ππ₯+π_2 cosβ‘ππ₯ )βπ^2 π^ππ (π_π πππβ‘ππ+π_π πππβ‘ππ ) Putting π¦ =π^ππ₯ (π1 πππ ππ₯+ π2 sinβ‘ππ₯) π¦^β²β²βππ¦^β²=πππ^ππ (γβπγ_π πππβ‘ππ+π_π πππβ‘ππ )βπ^2 π¦ "Putting" π¦^β²βππ¦=ππ^ππ₯ (γβπγ_1 π ππβ‘ππ₯+π_2 πππ β‘ππ₯ ) π¦^β²β²βππ¦^β²=π(π¦^β²βππ¦)βπ^2 π¦ π¦^β²β²βππ¦^β²=ππ¦^β²βπ^2 π¦βπ^2 π¦ π¦^β²β²βππ¦^β²βππ¦^β²+π^2 π¦+π^2 π¦=0 π^β²β²βπππ^β²+(π^π+π^π)π=π Hence verified