Ordinary Differential Equations Questions for Campus Interviews

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

Answer: c
Explanation: This is the special case of Legendre’s Function.
Assume log(2x+3)=z
Then, e^z=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 e^z)
=-4e^z-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, e^z=(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, e^z=(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, e^z=(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, e^z=(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) \(c_1(3x+5)^{0.76}+ c_2(3x+5)^{0.073}+\frac{-(15(cos⁡(log⁡(3x+5))+17 sin⁡(log⁡(3x+5)))))}{16} \)
View Answer

Answer: a
Explanation: The given equation is a Legendre’s Function.
Assume log⁡(3x+5)=z
Then, e^z=(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, e^z=(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, e^z=(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 =\(\frac{1}{4}×(-sin⁡(log⁡(2x+4))-cos⁡(log⁡(2x+4)))\).

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