CBSE Class 12 Sample Paper for 2018 Boards
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CBSE Class 12 Sample Paper for 2018 Boards
Last updated at December 16, 2024 by Teachoo
This is a question of CBSE Sample Paper - Class 12 - 2017/18.
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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)β+ (by2)β = (1)β 2ax + 2by ππ¦/ππ₯ = 0 2ax + 2byy1 = 0 2(ax + byy1) = 0 ax + byy1 = 0 Now, finding y2 From (1) ax + byy1 = 0 ax + byππ¦/ππ₯ = 0 Differentiating w.r.t. x (ax)β + ("by" ππ¦/ππ₯)^β²= 0 a + b("y" ππ¦/ππ₯)^β²= 0 a + b(π¦^β² ππ¦/ππ₯+π¦π¦β²β²)= 0 a + b(π¦^β² π¦β²+π¦π¦β²β²)= 0 a + b(π¦1 π¦1+π¦π¦2)= 0 a + b(π¦1 π¦1+π¦π¦2)= 0 a + b(γπ¦_1γ^2+π¦π¦2)= 0 a = β b(γπ¦_1γ^2+π¦π¦2) Now, from (1) ax + byy1 = 0 Putting a = β b(γπ¦_1γ^2+π¦π¦2) from (2) β b(γπ¦_1γ^2+π¦π¦2)x + byy1 = 0 byy1 = b(γπ¦_1γ^2+π¦π¦2)x Cancelling b both sides yy1 = (γπ¦_1γ^2+π¦π¦2)x x(γπ¦_1γ^2+π¦π¦2) = yy1 Hence proved
