The homogeneous linear ode for the given basis of solutions 1 and lnx is
(a) y"+y'=0
(b) y"=0
(c) x²y"=0
(d) xy"+y'=0
Answers
Answer:
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Class 12
>>Maths
>>Differential Equations
>>Solving Linear Differential Equation
>>The function y = f(x) is the solution of
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The function y=f(x) is the solution of the differential equation
dx
dy
+
x
2
−1
xy
=
1−x
2
x
4
+2x
in (−1,1) satisfying f(0)=0. Then ∫
−
2
3
2
3
f(x)dx is
Hard
JEE Advanced
Solution
verified
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Correct option is B)
dx
dy
+
x
2
−1
x
y=
1−x
2
x
4
+2x
This is a linear differential equation
I.F. =e
∫
x
2
−1
x
dx
=e
2
1
ln∣x
2
−1∣
=
1−x
2
⇒ solution is
y
1−x
2
=∫
1−x
2
x(x
3
+2)
⋅
1−x
2
dx
or y
1−x
2
=∫(x
4
+2x)dx=
x
5
5+x
2
+c
f(0)=0⇒c=0
⇒f(x)
1−x
2
=
5
x
5
+x
2
Now,
∫
−
3
/2
3
/2
f(x)dx=∫
3
/2
3
/2
d
1−x
2
x
2
dx (Using property)
=2∫
0
3
/2
1−x
2
x
2
dx=2∫
0
π/3
cosθ
sin
2
θ
cosθdθ (Taking x=sinθ)
=2∫
0
π/3
sin
2
θdθ=2[
2
θ
−
4
sin2θ
]
0
3
π
=2(
6
π
)−2(
8
3
)=
3
π
−
4
3