MCQ

MCQ 12 Marks

The order and degree of the differential equation $\frac{d^2 y}{d x^2}+\frac{d y}{d x}+x=\sqrt{1+\frac{d^3 y}{d x^3}}$ are respectively____.

A
2,3
B
3,2
C
1,3
D
3,1

Answer

3, 1

$
\frac{d^2 y}{d x^2}+\frac{d y}{d x}+x=\sqrt{1+\frac{d^3 y}{d x^3}}
$
Squaring both sides
$
\left(\frac{d^2 y}{d x^2}+\frac{d y}{d x}+x\right)^2=1+\frac{d^3 y}{d x^3}
$
$\therefore$ Order $=3$, Degree $=1$

View full question & answer→

MCQ 22 Marks

The solution of the differential equation $\frac{d y}{d x}=\sec x-y \tan x$ is .....

✓
$y \sec x=\tan x+ c$
B
$y \sec x+\tan x=c$
C
$\sec x=y \tan x+c$
D
$\sec x+y \tan x=c$

Answer

Correct option: A.

$y \sec x=\tan x+ c$

(A)
$\frac{d y}{d x}+y \tan x=\sec x$
The given equation is of the form
$\frac{d y}{d x}+P y=Q$
I. $f==e^{\int P d x}=e^{\int \tan x d x}$
$\begin{aligned} & =e^{\log |\sec x|} \\ & =\sec x\end{aligned}$
Solution of the given equation is
$\begin{aligned} & y \cdot(I \cdot F)=\int Q(I \cdot F) d x+c \\ & y \sec x=\int \sec x \sec x d x+c \\ & y \sec x=\tan x+c\end{aligned}$

View full question & answer→

MCQ 32 Marks

The differential equation of the family of curves $y=c_1 e^x+c_2 e^{-x}$ is .....

A
$\frac{d^2 y}{d x^2}+y=0$
B
$\frac{d^2 y}{d x^2}-y=0$
C
$\frac{d^2 y}{d x^2}+1=0$
D
$\frac{d^2 y}{d x^2}-1=0$

Answer

coming soon

View full question & answer→

MCQ 42 Marks

Order and degree of differential equation $\frac{d^4 y}{d x^4}=\left[1+\left(\frac{d y}{d x}\right)^2\right]^3$ respectively are ______.

A
Order : 1, Degree : 4
B
Order : 4, Degree : 1
C
Order : 6, Degree : 1
D
Order : 1, Degree : 6

Answer

coming soon

View full question & answer→

MCQ 52 Marks

If $y=a e^{5 x}+b e^{-6 x}$, then the differential equation is $\qquad$

A
$\frac{d^2 y}{d x^2}=25 y$
B
$\frac{d^2 y}{d x^2}=-25 y$
C
$\frac{d^2 y}{d x^2}=-5 y$
D
$\frac{d^2 y}{d x^2}=5 y$

Answer

coming soon

View full question & answer→

MCQ 62 Marks

Integrating factor of linear differential equation $x \frac{d y}{d x}+2 y=x^2 \log x$ is .....

A
$\frac{1}{x^2}$
B
$\frac{1}{x}$
C
$x$
D
$x^2$

Answer

$\begin{aligned} & x^2 \\ & \frac{d y}{d x}+\frac{2 y}{x}=x \log x \\ & P=\frac{2}{x} \\ & \text { I. } F=e^{\int \frac{2}{x} d x}=e^{2 \log x}=x^2\end{aligned}$

View full question & answer→

MCQ 72 Marks

The differential equation of $y=\frac{c}{x}+c^2$ is

A
$x^4\left(\frac{d y}{d x}\right)^2-x\frac {d y}{d x}=y$
B
$\frac{d^2 y}{d x^3}+x \frac{d y}{d x}+y=0$
C
$x^3\left(\frac{d y}{d x}\right)^2+x\frac{d y}{d x}=y$
D
$\frac{d^2 y}{d x^2}+\frac{d y}{d x}-y=0$

Answer

$y=\frac{c}{x}+c^2 ....(1)$
Differentiating w.r.t.x,
$\frac{d y}{d x}=\frac{-c}{x^2}+0$
$c=-x^2 \frac{ dy }{ dx }$......(2)
Putting in equation (1)
$y=\frac{-x^2 \frac{d y}{d x}}{x}+\left(-x^2 \frac{d y}{d x}\right)^2$
$y=-x \frac{d y}{d x}+x^4\left(\frac{d y}{d x}\right)^2$
$x^4\left(\frac{d y}{d x}\right)^2-x \frac{d y}{d x}=y$

View full question & answer→

MCQ 82 Marks

The equation of tangent to the curve $y=3 x^2-x+1$ at the point $(1,3)$ is

A
$y=5 x+2$
✓
$y=5 x-2$
C
$y=\frac{1}{5} x+2$
D
$y=\frac{1}{5} x-2$

Answer

Correct option: B.

$y=5 x-2$

$ \frac{d y}{d x}=6 x-1 \text { at }(1,3)$
Slope of the tangent at $(1, 3) = (6 - 1) = 5$
Equation of tangent is $y - y_1 = m(x - x_1)$
$y - 3 = 5(x - 1)$
$5x - y - 2 = 0$
$y = 5x - 2$

View full question & answer→

MCQ 92 Marks

Order and degree of the differential equation $\left[1+\left(\frac{d y}{d x}\right)^3\right]^{\frac{7}{3}}=7 \frac{d^2 y}{d x^2}$ are respectively....

A
2,3
B
3,2
C
7,2
D
3,7

Answer

coming soon

View full question & answer→

MCQ 102 Marks

Equation of the tangent to the curve $2 x^2+3 y^2-5=0$ at $(1,1)$ is ______.

A
$2 x-3 y-5=0$
B
$2 x+3 y-5=0$
C
$2 x+3 y+5=0$
D
$3 x+2 y+5=0$

Answer

coming soon

View full question & answer→

MCQ 112 Marks

The order and the degree of the differential equation $\left(\frac{d^3 y}{d x^3}\right)^{\frac{1}{3}}-\left(\frac{d u}{d x}\right)^{\frac{1}{3}}=0$ are respectively....

A
3,2
B
2,3
C
6,3
D
3,1

Answer

coming soon

View full question & answer→

MCQ 122 Marks

The Order and degree of the differential equation $\left[1+\left(\frac{d y}{d x}\right)^3\right]^{\frac{7}{3}}=7 \frac{d^2 y}{d x^2}$ are respectively....

A
2,3
B
3,2
C
7,2
D
3,7

Answer

$\left[1+\left(\frac{d y}{d x}\right)^3\right]^{\frac{7}{3}}=7 \frac{d^2 y}{d x^2}$
Cubing on both sides, we get
$\left[1+\left(\frac{d y}{d x}\right)^3\right]^7=7\left(\frac{d^2 y}{d x^2}\right)^3$
By definition of degree and order Degree: 3 ; Order: 2

View full question & answer→

Maths MCQ

Take a timed test