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12th Business Maths Weekly Test - 4 ( Differential Equations )

St. Britto Hr. Sec. School - Madurai

12th Business Maths Weekly Test - 4 ( Differential Equations )-Aug 2020

Time : 12.45 Hours

Part - A

Multiple Choice Question - Answer All Question

The general solution of the differential equation \(\frac{dy}{dx}\) = cos x is

y = sinx + 1
y = sinx - 2
y = cos x + c, c is an arbitrary constant
y = sin x + c, c is an arbitrary constant

If y=cx + c− c³ then its differential equation is

\(y=x\frac { dy }{ dx } +\frac { dy }{ dx } -{ \left( \frac { dy }{ dx } \right) }^{ 3 }\)
\(y={ \left( \frac { dy }{ dx } \right) }^{ 3 }=x\frac { dy }{ dx } -\frac { dy }{ dx } \)
\(\frac { dy }{ dx } +y={ \left( \frac { dy }{ dx } \right) }^{ 3 }-x\frac { dy }{ dx } \)
\(\frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } =0\)

A homogeneous differential equation of the form \(\frac { dy }{ dx } \) = f\(\left( \frac { x }{ y } \right) \) can be solved by making substitution,

x = v y
y = v x
y = v
x = v

The differential equation \({ \left( \frac { dx }{ dy } \right) }^{ 3 }+2{ y }^{ \frac { 1 }{ 2 } }\) = x is

of order 2 and degree 1
of order 1 and degree 3
of order 1 and degree 6
of order 1 and degree 2

Which of the following is the homogeneous differential equation?

(3x−5)dx = (4y−1)dy
xy dx−(x³+y³)dy = 0
y²dx+(x²− xy − y²)dy = 0
(x²+y)dx = (y²+x)dy

The solution of the differential equation \(\frac { dy }{ dx } =\frac { y }{ x } +\frac { f\left( \frac { y }{ x } \right) }{ f'\left( \frac { y }{ x } \right) } \) is

\(f\left( \frac { y }{ x } \right) =kx\)
\(xf\left( \frac { y }{ x } \right) =k\)
\(f\left( \frac { y }{ x } \right) =ky\)
\(yf\left( \frac { y }{ x } \right) =k\)

A homogeneous differential equation of the form \(\frac { dy }{ dx } \) = f\(\left( \frac { y }{ x } \right) \) can be solved by making substitution,

y = v x
v = y x
x = v y
x = v

The degree of the differential equation \(\frac { { d }^{ 4 }y }{ { dx }^{ 4 } } { -\left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) }^{ 4 }+\frac { dy }{ dx } =3\)

1
2
3
4

The differential equation of y = mx + c is (m and c are arbitrary constants)

\(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \) = 0
y = x \(\frac { dy }{ dx } \) + c
xdy + ydx = 0
ydx − xdy = 0

The complementary function of \(\frac { { d }^{ 3 }y }{ d{ x }^{ 3 } } -\frac { dy }{ dx } \) = 0 is

A + Be^x
(A + B)e^x
(Ax + B)e^x
Ae^x + B

The integrating factor of the differential equation \(\frac{dx}{dy}+Px=Q\)

e^ഽPdx
e^ഽPdx
ഽPdy
e^ഽPdy

The order and degree of the differential equation \({ \left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) }^{ \frac { 1 }{ 2 } }-\sqrt { \frac { dy }{ dx } } -4=0\) are respectively

2 and 6
3 and 6
1 and 4
2 and 4

The complementary function of (D²+ 4)y = e^2x is

(Ax +B)e^2x
(Ax +B)e^−2x
A cos 2x + B sin 2x
Ae^−2x+ Be^2x

The differential equation formed by eliminating a and b from y=ae^x+ be^−x is

\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -y=0\)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -\frac { dx }{ dy } =0\)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =0\)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -x=0\)

The integrating factor of x \(\frac { dy }{ dx } \) - y = x² is

\(\frac {-1}{x}\)
\(\frac {1}{x}\)
log x
x

The P.I of (3D²+ D − 14)y = 13e^2x is

\(\frac {x}{2}\)e^2x
xe^2x
\(\frac {x^2}{2}\)e^2x
13xe^2x

If sec² x is an integrating factor of the differential equation \(\frac { dy }{ dx } \) + Py Q then P =

2 tan x
sec x
cos² x
tan² x

The differential equation formed by eliminating A and B from y = e^−2x(A cos x + B sin x) is

y₂− 4y₁+ 5 = 0
y₂+ 4y – 5 = 0
y₂−4y₁−5= 0
y₂+ 4y₁+ 5 = 0

The differential equation of x²+ y²= a²

xdy + ydx = 0
ydx – xdy = 0
xdx – ydx = 0
xdx + ydy = 0

The particular integral of the differential equation f(D)y = e^ax where f(D) = (D−a)²

\(\frac { { x }^{ 2 } }{ 2 } { e }^{ ax }\)
xe^ax
\(\frac { x }{ 2 } \)e^ax
x²e^ax

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