# 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) $$c_1(\frac{1}{(8x+7)})^\frac{1}{4}+\frac{(8x+7)}{80}-\frac{7}{16}$$

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) $$c_1(2x+3)^\frac{3}{2}, -4(2x+3)-4$$

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

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)×$$\frac{cos⁡(log⁡(x+1))}{2}$$

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

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) $$c_1(3x+2)^{0.022}+(3x+2)^{1.15}(c_2 sin⁡(0.22 log⁡(3x+2))+c_3 sin(0.22 log⁡(3x+2)))$$

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

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

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

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

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