Verify that ax 2 + by 2 = 1 is a solution of the differential equation x(yy 2 + y 1 2 ) = yy 1

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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)โ€™+ (by2)โ€™ = (1)โ€™ 2ax + 2by ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 0 2ax + 2byy1 = 0 2(ax + byy1) = 0 ax + byy1 = 0 Now, finding y2 From (1) ax + byy1 = 0 ax + by๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 0 Differentiating w.r.t. x (ax)โ€™ + ("by" ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ)^โ€ฒ= 0 a + b("y" ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ)^โ€ฒ= 0 a + b(๐‘ฆ^โ€ฒ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ+๐‘ฆ๐‘ฆโ€ฒโ€ฒ)= 0 a + b(๐‘ฆ^โ€ฒ ๐‘ฆโ€ฒ+๐‘ฆ๐‘ฆโ€ฒโ€ฒ)= 0 a + b(๐‘ฆ1 ๐‘ฆ1+๐‘ฆ๐‘ฆ2)= 0 a + b(๐‘ฆ1 ๐‘ฆ1+๐‘ฆ๐‘ฆ2)= 0 a + b(ใ€–๐‘ฆ_1ใ€—^2+๐‘ฆ๐‘ฆ2)= 0 a = โ€“ b(ใ€–๐‘ฆ_1ใ€—^2+๐‘ฆ๐‘ฆ2) Now, from (1) ax + byy1 = 0 Putting a = โ€“ b(ใ€–๐‘ฆ_1ใ€—^2+๐‘ฆ๐‘ฆ2) from (2) โ€“ b(ใ€–๐‘ฆ_1ใ€—^2+๐‘ฆ๐‘ฆ2)x + byy1 = 0 byy1 = b(ใ€–๐‘ฆ_1ใ€—^2+๐‘ฆ๐‘ฆ2)x Cancelling b both sides yy1 = (ใ€–๐‘ฆ_1ใ€—^2+๐‘ฆ๐‘ฆ2)x x(ใ€–๐‘ฆ_1ใ€—^2+๐‘ฆ๐‘ฆ2) = yy1 Hence proved

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