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
Made by
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.
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