Differential Equations Test

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Differential Equations Test - 3

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Question 1
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Write the order and degree of the given differential equation:

A
2,3

B
1,1

C
2,1

D
3, 1

Solution

Order = 3 and Degree = 1
The order of a differential equation is determined by the highest-order derivative; the degree is determined by the highest power on a variable. The higher the order of the differential equation, the more arbitrary constants need to be added to the general solution.
Question 2
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The order of a differential equation representing a family of curves is same as:

A
The number of arbitrary constants present in the equation.

B
The number of variables in the equation

C
The degree of the equation

D
The number of curves in the family

Solution

The order of a differential equation representing a family of curves is same as the number of arbitrary constants present in the equation corresponding to the family of curves.
Question 3
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The solution of the equation x³ dx + (y + 1)² dy = 0

A
12[x⁴ - (y +1)³ ] = c

B
12[x⁴ + (y +1)³ ] = c

C
3x⁴ + 4(y +1)³ = c

D
3x⁴ + 3(y +1)³ = c

Solution

x³ dx + (y + 1)dy = 0
=>d(x⁴ /4) + d((y + 1)³ /3) = 0
=>d(x⁴ /4 + ((y + 1)³ /3)) = 0
=>(x⁴ /4 + ((y + 1)³ /3)) = constt
=>12 *(x⁴ /4 + ((y + 1)³ /3)) = 12 * constt
=>3x⁴ + 4(y + 1)³ = constt
Question 4
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Identify the type of differential equation

A
Variable separable

B
Homogeneous

C
Linear

D
Both Variable separable and Linear

Solution

(x −y)dy/dx = x + 2y
⇒dy/dx = (x + 2y)/(x −y)
F (x,y) = (x + 2y)/(x −y)
F(Ax, Ay) = (Ax + 2Ay)/(Ax −Ay)
= A(x + 2y)/A(x −y)
= (x + 2y)/(x −y) = F (x,y)
Hence, the equation is homogenous.
Question 5
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Identify the form of the given Differential Equation

A
Variable separable

B
Homogeneous

C
Linear

D
None of the above
Question 6
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The solution of the initial value problem is :

A
sec x = 2y

B
x = 2cosy

C
y = 2secx

D
y = 2cosx

Solution

xdy/dx = -coty
dy/dx = - coty/x
dy/coty = - dx/x
tany dy = - dx/x
∫(siny/cosy)dy = - ∫dx/x …………..(1)
Put t = cosy
dt = -sinydy
Put the value of dt in eq(1)
-∫dt/t = -∫dx/x
log |t| = log |x| + log c
log |cosy| = log(c.|x|)
|cos y| = c.|x|....................(2)
As y = π/4, x = (2)^½
cos π/4 = c*(2)^½
1/(2)^½ = c*(2)^½
c = ½
Put the value of ‘c ’in eq(2)
cos y = x/2 =>x = 2cosy
Question 7
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The solution of the differential equation is :

A
a log sec by + be^ax = c

B
a log (sec by + tan by) = be^ax = c

C
a log (sec by + tan by) + be^ax = c

D
a log sec by + be^ax + c

Solution

dy/dx = e^ax cos by
∫dy/cos by = ∫e^ax dx
∫sec by dy = ∫e^ax dx
= (log| sec by + tan by|)/b = e^ax /a + c
= a(log| sec by + tan by|) = be^ax + c
Question 8
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The solution of the differential equation is :

A
2y = x² (log x + 1) + c

B

C

D
y = x² (log x + 1) + c

Solution

dy/dx = x logx
=>∫dy = ∫x logx dx
y = logx . x² /2 - ∫1/x . x² /2 dx + c
y = x² /2 log x -½∫x dx + c
y = x² /2 log x - ½. x² /2 + c
y = x² /2 log x - x² /4 + c
Question 9
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The solution of the differential equation is :

A
y² = 2xe^x + 2e^x + cx + d

B
y = 2xe^x + 2e^x + cx + d

C
y = xe^x - 2e^x + cx + d

D
y = 2xe^x - 2e^x + cx + d

Solution

d² y/dx² = xe^x
d² y/dx² = ∫xe^x
d/dx(dy/dx) = [xe^x ]
Integrating, we get
dy/dx = (x-1)e^x + c1
Again integrating, y = (x-2)e^x + c1x + c2
Question 10
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The solution of the initial value problem e^dy/dx = x + 1, y(0) = 3 is :

A
y = (x –1)log(x + 1) + 3

B
y = (x + 1)log(x + 1) -x+3

C
y = (x + 1)log(x + 1) + 4

D
y = (x + 1)log(x + 1) –1

Solution

e^dy/dx = x + 1
⇒dy/dx = log(x+1)
⇒dy = log(x+1)dx
Integrating both sides, we get
⇒∫dy = ∫log(x+1)dx
y = log(x + 1) ∫1dx - ∫[d/dx{log(x + 1)} ∫1dx]dx
y = xlog(x + 1) - ∫1/(x+1)xdx
y = xlog(x + 1) - ∫(1 - 1/(x + 1)dx
y = xlog(x + 1) - ∫dx + ∫1/(x+1)dx
y = xlog(x + 1) - x + log|x + 1| + c
y = (x + 1)log|x + 1| - x + c
Here c is 3
y = (x + 1)log|x + 1| - x + 3