Ordinary Differential Equations Questions and Answers – Special Functions – 4 (Legendre)

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This set of 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) \(c_1(\frac{1}{(8x+7)})^\frac{1}{4}+\frac{(8x+7)}{80}-\frac{7}{16}\)
View Answer

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 = \(c_1(\frac{1}{(8x+7)})^\frac{1}{4}+\frac{(8x+7)}{80}-\frac{7}{16}.\)
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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) \(c_1(2x+3)^\frac{3}{2}, -4(2x+3)-4\)
View Answer

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) \(c_1(2x+3)+ c_2(2x+3)^\frac{-1}{2}\)
View Answer

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)×\(\frac{cos⁡(log⁡(x+1))}{2}\)
View Answer

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) \(\frac{2x}{3}+4+\frac{1}{(x+3)}×c_1cos⁡(\frac{\sqrt{3}}{2} log⁡(x+3))+c_2sin⁡(\frac{\sqrt{3}}{2} log⁡(x+3))\)
View Answer

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
\(\frac{1}{(x+3)}× c_1cos⁡(\frac{\sqrt{3}}{2} log⁡(x+3))+c_2sin⁡(\frac{\sqrt{3}}{2} log⁡(x+3))+\frac{4x}{3}+4\)
.
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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) \(c_1(3x+2)^{0.022}+(3x+2)^{1.15}(c_2 sin⁡(0.22 log⁡(3x+2))+c_3 sin(0.22 log⁡(3x+2)))\)
View Answer

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=\( c_1(3x+2)^{0.026}+(3x+2)^{1.15} (c_2cos⁡(0.22 log⁡(3x+2))+c_3sin(0.22 log⁡(3x+2)))\).

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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) \(c_1(3x+5)^{0.76}+ c_2(3x+5)^{0.073}+\frac{-(15(cos⁡(log⁡(3x+5))+17 sin⁡(log⁡(3x+5)))))}{16} \)
View Answer

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) \(\frac{1}{(x+2)}×(c_1cos⁡(log⁡(x+2))+c_2sin⁡(log⁡(x+2)))\)
View Answer

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) \(\frac{1}{2}×(-sin⁡(log⁡(2x+4))+cos⁡(log⁡(2x+4)))\)
View Answer

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 =\(\frac{1}{4}×(-sin⁡(log⁡(2x+4))-cos⁡(log⁡(2x+4)))\).

Sanfoundry Global Education & Learning Series – Ordinary Differential Equations.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn