Question 10
Last updated at Dec. 21, 2017 by Teachoo

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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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)โ+ (by2)โ = (1)โ
2ax + 2by ๐๐ฆ/๐๐ฅ = 0
2ax + 2byy1 = 0
2(ax + byy1) = 0
ax + byy1 = 0
Now, finding y2
From (1)
ax + byy1 = 0
ax + by๐๐ฆ/๐๐ฅ = 0
Differentiating w.r.t. x
(ax)โ + ("by" ๐๐ฆ/๐๐ฅ)^โฒ= 0
a + b("y" ๐๐ฆ/๐๐ฅ)^โฒ= 0
a + b(๐ฆ^โฒ ๐๐ฆ/๐๐ฅ+๐ฆ๐ฆโฒโฒ)= 0
a + b(๐ฆ^โฒ ๐ฆโฒ+๐ฆ๐ฆโฒโฒ)= 0
a + b(๐ฆ1 ๐ฆ1+๐ฆ๐ฆ2)= 0
a + b(๐ฆ1 ๐ฆ1+๐ฆ๐ฆ2)= 0
a + b(ใ๐ฆ_1ใ^2+๐ฆ๐ฆ2)= 0
a = โ b(ใ๐ฆ_1ใ^2+๐ฆ๐ฆ2)
Now, from (1)
ax + byy1 = 0
Putting a = โ b(ใ๐ฆ_1ใ^2+๐ฆ๐ฆ2) from (2)
โ b(ใ๐ฆ_1ใ^2+๐ฆ๐ฆ2)x + byy1 = 0
byy1 = b(ใ๐ฆ_1ใ^2+๐ฆ๐ฆ2)x
Cancelling b both sides
yy1 = (ใ๐ฆ_1ใ^2+๐ฆ๐ฆ2)x
x(ใ๐ฆ_1ใ^2+๐ฆ๐ฆ2) = yy1
Hence proved

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