

Get live Maths 1-on-1 Classs - Class 6 to 12
Examples
Example 1 (ii) Important
Example 1 (iii) Important
Example 2
Example 3 Important
Example 4 Deleted for CBSE Board 2023 Exams
Example 5 Deleted for CBSE Board 2023 Exams
Example 6 Important Deleted for CBSE Board 2023 Exams
Example 7 Deleted for CBSE Board 2023 Exams
Example 8 Deleted for CBSE Board 2023 Exams
Example 9
Example 10
Example 11
Example 12 Important
Example 13
Example 14 Important
Example 15 Important
Example 16
Example 17 Important
Example 18 Important
Example 19
Example 20 Important
Example 21
Example 22 Important
Example 23 Important
Example 24 You are here
Example 25 Deleted for CBSE Board 2023 Exams
Example 26
Example 27 Important
Example 28 Important
Last updated at March 30, 2023 by Teachoo
Example 24 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₯ (π1 πππ ππ₯+ π2 sinβ‘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_1 sinβ‘ππ₯+π_2 cosβ‘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