Ordinary Differential Equations Questions and Answers for Campus interviews focuses on “Special Functions – 4 (Legendre)”.
1. Find the General Solution of the given differential equation.
((8x+7)frac{dy}{dx}+2y=x)
a) (c_1(frac{1}{(8x+7)})^frac{1}{4}+frac{(8x+7)}{40}-frac{5}{16})
b) (c_1(frac{1}{(8x+7)})^frac{1}{4}+frac{(8x+7)}{80}-frac{7}{8})
c) (c_1(frac{1}{(8x+7)})^frac{1}{4}+frac{(8x+7)}{40}-frac{7}{16})
d) (c_1(frac{1}{(8x+7)})^frac{1}{4}+frac{(8x+7)}{80}-frac{7}{16})
Answer: d
Explanation: This special case where D has a co-efficient is solved using Legendre’s method.
Let log(8x+7)=z
Then, ez=8x+7
((8x+7)frac{dy}{dx}=8Dy), where D=(frac{d}{dz} )
Substituting this in the equation,
8Dy+2y=x
y(8D+2)=x
Thus the auxiliary equation is 8D+2=0
Thus, D = (frac{-1}{4})
C.F = (c_1 e^{frac{-z}{4}} = c_1(frac{1}{(8x+7)})^frac{1}{4})
P.I = (frac{1}{(8D+2)} × x)
= (frac{1}{(8D+2)} × (frac{e^z-7}{8}))
= (frac{1}{(8D+2)} × frac{e^z}{8} – frac{1}{(8D+2)} × frac{7}{8} × e^0 )(Solving by substituting powers of ez)
= (frac{e^z}{80}-frac{7}{16})
=(frac{(8x+7)}{80}-frac{7}{16})
The General solution is C.F+P.I = (c_1(frac{1}{(8x+7)})^frac{1}{4}+frac{(8x+7)}{80}-frac{7}{16}.)
2. Find the Continuous Function and Particular Integral for the given differential equation.
((2x+3)frac{dy}{dx}-3y=8x)
a) (c_1(2x+3)^frac{5}{2}, -4(2x+3)-4)
b) (c_1(2x+3)^frac{3}{2}, -2(2x+3)-4)
c) (c_1(2x+3)^frac{3}{2}, -4(2x+3)-4)
d) (c_1(2x+3)^frac{3}{2}, -4(2x+3)-4)
Answer: c
Explanation: This is the special case of Legendre’s Function.
Assume log(2x+3)=z
Then, ez=2x+3
((2x+3)frac{dy}{dx}=2Dy), where D=(frac{d}{dz})
Substituting this in the equation,
2Dy-3y=8x
y(2D-3)=8x
Thus the auxiliary equation is 2D-3=0
Thus, D=(frac{3}{2})
C.F=(c_1 e^frac{3z}{2} = c_1(2x+3)^frac{3}{2})
P.I=(frac{1}{(2D-3)}×8x)
=(frac{1}{(2D-3)}×8(frac{e^z-3}{2}))
=(frac{1}{(2D-3)}×e^frac{z}{2}×8-frac{1}{(2D-3)}×8×frac{-3}{2}×e^0) (Solving by substituting powers of ez)
=-4ez-4
=-4(2x+3)-4
Thus, C.F is (c_1(2x+3)^frac{3}{2})
And P.I is -4(2x+3)-4.
3. Find the C.F of the following Differential Equation.
((2x+3)^2 frac{d^2 y}{dx^2} + (2x+3)frac{dy}{dx} – 2y = x )
a) (c_1(2x+3)+ c_2(2x+3)^frac{-1}{2})
b) (c_1(2x+3)+ c_2(2x+3)^frac{1}{2})
c) (c_1(2x+3)+ c_2(2x+3)^frac{-3}{2})
d) (c_1(2x+3)+ c_2(2x+3)^frac{-1}{2})
Answer: a
Explanation: Assume log(2x+3)=z
Then, ez=(2x+3)
(2x+3)(frac{dy}{dx})=2Dy, Where D is (frac{d}{dz})
((2x+3)^2 frac{d^2 y}{dx^2}=2^2 D(D-1)y)
Substituting in the equation given
4D(D-1)y+2Dy-2y=x
y(4D(D-1)+2D-2)=x
y(4D2-2D-2)=x
Thus, the Auxiliary Equation is 4D2-2D-2
D=1 or D=(frac{-1}{2})
Thus, the C.F for the given equation is (c_1e^z + c_2e^frac{-z}{2}=c_1(2x+3) + c_2(2x+3)^frac{-1}{2}).
4. Find the P.I of the given Differential Equation.
((x+1)^2 frac{d^2 y}{dx^2}+(x+1) frac{dy}{dx}+y=sin(log(1+x)))
a) -log(x+1)×(frac{cos(log(x+1))}{2})
b) -log(x+1)×(frac{sin(log(x+1))}{2})
c) log(x+1)×(frac{cos(log(x+1))}{2})
d) -log(x+2)×(frac{cos(log(x+1))}{2})
Answer: a
Explanation: Assume log(x+1)=z
Then, ez=(x+1)
((x+1)frac{dy}{dx}=Dy), Where D is (frac{d}{dz})
((x+1)^2 frac{d^2 y}{dx^2}=D(D-1)y)
Substituting in the equation given
D(D-1)y+Dy+y = sin(log(1+x))
y(D(D-1)+D+1) = sin(log(1+x))
y(D2+1) = sin(log(1+x))
Thus, the Auxiliary Equation is D2+1
P.I = (frac{1}{D^2+1})×sin(z)
To find P.I, substitute D2=-(1)2
Since the denominator becomes zero, multiply the numerator by z and differentiate the denominator.
P.I = (frac{z}{2D})×sin(z)
=(frac{-zcos(z)}{2})
=-log(x+1)×(frac{cos(log(x+1))}{2}).
5. Solve this Differential Equation to find its General Solution.
((x+3)frac{d^2y}{dx^2}+2 frac{dy}{dx}+frac{y}{(x+3)}=4)
a) (frac{4x}{3}+2+frac{1}{(x+3)}×c_1cos(frac{sqrt{3}}{2} log(x+3))+c_2sin(frac{sqrt{3}}{2} log(x+3)))
b) (frac{4x}{3}+4+frac{1}{(x+3)}×c_1cos(frac{sqrt{3}}{2} log(x+3))+c_2sin(frac{sqrt{3}}{2} log(x+3)))
c) (x+4+frac{1}{(x+3)}×c_1cos(frac{sqrt{3}}{2} log(x+3))+c_2sin(frac{sqrt{3}}{2} log(x+3)))
d) (frac{2x}{3}+4+frac{1}{(x+3)}×c_1cos(frac{sqrt{3}}{2} log(x+3))+c_2sin(frac{sqrt{3}}{2} log(x+3)))
Answer: b
Explanation: Multiply both the sides with (x+3)
We get the equation,
((x+3)^2 frac{d^2 y}{dx^2}+2(x+3) frac{dy}{dx}+y=4(x+3))
This equation is in Legendre’s form.
Assume log(x+3)=z
Then, ez=(x+3)
((x+3)frac{dy}{dx}=Dy), Where D is (frac{d}{dz})
((x+3)^2 frac{d^2 y}{dx^2}=D(D-1)y)
Substituting in the equation given
D(D-1)y+2Dy+y=4(x+3)
y(D(D-1)+2D+1)=4(x+3)
y(D2+D+1)=4(x+3)
Thus, the Auxiliary Equation is D2+D+1=0
(D=frac{-1}{2}+frac{sqrt{3}}{2}i ,or. D=frac{-1}{2}-frac{sqrt{3}}{2}i)
Thus, the C.F for the given equation is (e^{frac{-z}{2}} (c_1cos(frac{sqrt{3}}{2}z)+c_2sin(frac{sqrt{3}}{2}z)))
C.F = (frac{1}{(x+3)}× c_1cos(frac{sqrt{3}}{2} log(x+3))+c_2sin(frac{sqrt{3}}{2} log(x+3)))
P.I = (frac{1}{D^2+D+1}×4e^z)
To find P.I, substitute D=1
P.I = (frac{4×e^z}{3})
= (frac{4×(x+3)}{3})
= (frac{4x}{3}+4)
Thus, the general solution is
(frac{1}{(x+3)}× c_1cos(frac{sqrt{3}}{2} log(x+3))+c_2sin(frac{sqrt{3}}{2} log(x+3))+frac{4x}{3}+4)
.
6. Find the C.F for the following Differential Equation.
((3x+2)^3 frac{d^3y}{dx^3}+2(3x+2)^2 frac{d^2 y}{dx^2}+(3x+2) frac{dy}{dx}-y=(3x+2)^2)
a) (c_1(3x+2)^{0.022}+(3x+2)^{1.15}(c_2 cos(0.22 log(3x+2))+c_3 sin(0.22 log(3x+2))))
b) (c_1(3x+2)^{0.022}+(3x+2)^{1.15}(c_2 cos(0.22 log(3x+3))+c_3 sin(0.22 log(3x+2))))
c) (c_1(3x+2)^{0.026}+(3x+2)^{1.15}(c_2 cos(0.22 log(3x+2))+c_3 sin(0.22 log(3x+2))))
d) (c_1(3x+2)^{0.022}+(3x+2)^{1.15}(c_2 sin(0.22 log(3x+2))+c_3 sin(0.22 log(3x+2))))
Answer: c
Explanation: This equation is in the Legendre Form.
Assume log(3x+2)=z
Then, ez=(3x+2)
((3x+2) frac{dy}{dx}=3Dy), Where D is (frac{d}{dz})
((3x+2)^2 frac{d^2 y}{dx^2}=3^2 D(D-1)y)
((3x+2)^3 frac{d^3 y}{dx^3}=3^3 D(D-1)(D-2)y)
Substituting in the equation given
27D(D-1)(D-2)y+18D(D-1)y+3Dy-y=(3x+2)2
y(27(D(D-1)(D-2))+18(D(D-1))+3D-1)=(3x+2)2
y(27D3-63D2+39D-1)
Thus the Auxiliary Equation is 27D3-63D2+39D-1=0
D=0.026 or D=1.15+0.22i or D=1.15-0.22i
Thus, the C.F of the equation is given by
C.F=(c_1e^{0.026z}+e^{1.15z} (c_2cos(0.22z)+c_3sin(0.22z)))
C.F=( c_1(3x+2)^{0.026}+(3x+2)^{1.15} (c_2cos(0.22 log(3x+2))+c_3sin(0.22 log(3x+2)))).
7. Find the solution for the given Higher Order Differential Equation.
(2(3x+5)^2 frac{d^2 y}{dx^2}+(3x+5) frac{dy}{dx}+y=sin(log(3x+5)))
a) (c_1(3x+5)^{0.76}+ c_2(3x+5)^{0.073}+frac{-(15(cos(log(3x+5))+17 sin(log(3x+5)))))}{64})
b) (c_1(3x+5)^{0.76}+ c_2(3x+7)^{0.073}+frac{-(15(cos(log(3x+5))+17 sin(log(3x+5)))))}{64} )
c) (c_1(3x+5)^{0.76}+ c_2(3x+7)^{0.073}+frac{-(15(cos(log(3x+5))+17 sin(log(3x+5)))))}{16} )
d) (c_1(3x+5)^{0.76}+ c_2(3x+5)^{0.073}+frac{-(15(cos(log(3x+5))+17 sin(log(3x+5)))))}{16} )
Answer: a
Explanation: The given equation is a Legendre’s Function.
Assume log(3x+5)=z
Then, ez=(3x+5)
((3x+5) frac{dy}{dx}=3Dy), Where D is (frac{d}{dz})
((3x+5)^2 frac{d^2 y}{dx^2}=3^2 D(D-1)y)
Substituting in the equation given
(2×9)D(D-1)y+3Dy+y= sin(log(1+x))
y(18(D2-D)+3D+1) = sin(log(1+x))
y(18D2-15D+1) = sin(log(1+x))
Thus, the Auxiliary Equation is 18D2-15D+1
D=0.76 or D=-0.073
Thus, the C.F for the given equation is
(c_1e^{0.76}z+ c_2e^{0.073}z=c_1(3x+5)^{0.76}+ c_2(3x+5)^{0.073})
P.I = (frac{1}{(18D^2-15D+1)}×sin(z))
To find P.I, substitute D2=-(1)2
=(frac{1}{(-17-15D)}×sin(z))
Multiplying numerator and Denominator with factor(17-15D)
=(frac{(-15D+17)}{64}×sin(z)), Using a2-b2=(a+b)(a-b) and substituting D2=-(1)2
=((15(-cos(z)-17 sin(z)))/64)
=(frac{-(15(cos(log(3x+5))+17 sin(log(3x+5)))))}{64})
Thus, the general solution is
(c_1(3x+5)^{0.76}+ c_2(3x+5)^{0.073}+frac{-(15(cos(log(3x+5))+17 sin(log(3x+5)))))}{64})
.
8. Find the C.F for the given Higher Order Differential Equation.
(x^2 frac{d^2 y}{dx^2}+3(x+2) frac{dy}{dx}+4(1+x) frac{d^2 y}{dx^2}+2y=x)
a) (frac{1}{(x+1)}×(c_1cos(log(x+2))+c_2sin(log(x+2))))
b) (frac{1}{(x+2)}×(c_1cos(log(x+2))+c_2cos(log(x+2))))
c) (frac{1}{(x+1)}×(c_1cos(log(x+1))+c_2sin(log(x+1))))
d) (frac{1}{(x+2)}×(c_1cos(log(x+2))+c_2sin(log(x+2))))
Answer: d
Explanation: Bring all terms of (frac{d^2 y}{dx^2}) together.
((x+2)^2 frac{d^2 y}{dx^2}+3(x+2) frac{dy}{dx}+2y=x)
This is in Legendre’s Form.
Let, log(x+2)=z
Then, ez=(x+2)
((x+2) frac{dy}{dx}=Dy), Where D is (frac{d}{dz})
((x+2)^2 frac{d^2 y}{dx^2}=D(D-1)y)
Substituting in the equation given
D(D-1)y+3Dy+2y=x
y((D2-D)+3D+2)=x
y(D2+D+2)=x
Thus, the Auxiliary Equation is y(D2+D+2)=0
D=-1+i or D=-1-i
C.F is
(e^{-z}(c_1cos(z)+c_2sin(z))=frac{1}{(x+2)}×(c_1cos(log(x+2))+c_2sin(log(x+2)))).
9. Find the P.I for the given Differential Equation.
(16x^2 frac{d^2 y}{dx^2}+(2x+4) frac{dy}{dx}+4x(x+4) frac{d^2 y}{dx^2}+2y=cos(log(2x+4)))
a) (frac{1}{4}×(-sin(log(2x+4))+cos(log(2x+4))))
b) (frac{1}{4}×(-sin(log(2x+4))-cos(log(2x+4))))
c) (frac{1}{2}×(-sin(log(2x+4))-cos(log(2x+4))))
d) (frac{1}{2}×(-sin(log(2x+4))+cos(log(2x+4))))
Answer: b
Explanation: Bring all terms of ( frac{d^2 y}{ dx^2}) together.
The equation becomes-
((2x+4)^2 frac{d^2 y}{dx^2}+(2x+4) frac{dy}{dx}+2y= cos(log(2x+4)))
This is in Legendre’s Form.
Let, log(2x+4)=z
Then, ez=(2x+4)
((2x+4) frac{dy}{dx}=2Dy), Where D is (frac{d}{dz})
((2x+4)^2 frac{d^2 y}{dx^2}=4D(D-1)y)
Substituting in the equation given
4D(D-1)y+2Dy+2y=cos(log(2x+4))
y(4(D2-D)+2D+2)=cos(log(2x+4))
y(4D2-2D+2)=cos(log(2x+4))
Thus, the Auxiliary Equation is y(4D2-2D+2)=0
P.I = (frac{1}{4D^2-2D+2}×cos(log(2x+4)))
= (frac{1}{4D^2-2D+2}×cos(z))
In case of cos() function, Substitute D2=-(1)2
= (frac{1}{-2-2D}×cos(z)=frac{-1}{2}×frac{1}{(D+1)}×cos(z))
Multiplying numerator and Denominator with factor(D-1)
=(frac{1}{4}×(D-1)×cos(z))
=(frac{1}{4}×(-sin(z)-cos(z)))
=(frac{1}{4}×(-sin(log(2x+4))-cos(log(2x+4))))
Thus, the P.I for the given equation is =(frac{1}{4}×(-sin(log(2x+4))-cos(log(2x+4)))).
Global Education & Learning Series – Ordinary Differential Equations.
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