CBSE Class 12 Sample Paper for 2018 Boards
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CBSE Class 12 Sample Paper for 2018 Boards
Last updated at Dec. 16, 2024 by Teachoo
This is a question of CBSE Sample Paper - Class 12 - 2017/18.
You can download the question paper here https://www.teachoo.com/cbse/sample-papers/
Transcript
Question 10 Verify that ax2 + by2 = 1 is a solution of the differential equation x(yy2 + y12) = yy1 Given ax2 + by2 = 1 First we find y1, and y2 Now, ax2 + by2 = 1 Differentiating w.r.t. x (ax2)’+ (by2)’ = (1)’ 2ax + 2by 𝑑𝑦/𝑑𝑥 = 0 2ax + 2byy1 = 0 2(ax + byy1) = 0 ax + byy1 = 0 Now, finding y2 From (1) ax + byy1 = 0 ax + by𝑑𝑦/𝑑𝑥 = 0 Differentiating w.r.t. x (ax)’ + ("by" 𝑑𝑦/𝑑𝑥)^′= 0 a + b("y" 𝑑𝑦/𝑑𝑥)^′= 0 a + b(𝑦^′ 𝑑𝑦/𝑑𝑥+𝑦𝑦′′)= 0 a + b(𝑦^′ 𝑦′+𝑦𝑦′′)= 0 a + b(𝑦1 𝑦1+𝑦𝑦2)= 0 a + b(𝑦1 𝑦1+𝑦𝑦2)= 0 a + b(〖𝑦_1〗^2+𝑦𝑦2)= 0 a = – b(〖𝑦_1〗^2+𝑦𝑦2) Now, from (1) ax + byy1 = 0 Putting a = – b(〖𝑦_1〗^2+𝑦𝑦2) from (2) – b(〖𝑦_1〗^2+𝑦𝑦2)x + byy1 = 0 byy1 = b(〖𝑦_1〗^2+𝑦𝑦2)x Cancelling b both sides yy1 = (〖𝑦_1〗^2+𝑦𝑦2)x x(〖𝑦_1〗^2+𝑦𝑦2) = yy1 Hence proved
