Question 11 Mark
The general solution of the differential equation $\frac{y d x-x d y}{y}=0$ is
AnswerIt is given that $\frac{\mathrm{ydx}-\mathrm{xdy}}{\mathrm{y}}=0$
$\Rightarrow \frac{\mathrm{ydx}-\mathrm{xdy}}{\mathrm{xy}}=0$
$\Rightarrow \frac{1}{x} d x-\frac{1}{y} d y=0$
Integrating both sides, we get,
log|x| - log|y| = log k
$\Rightarrow \log \left|\frac{x}{y}\right|=\log k$
$\Rightarrow \frac{x}{y}=k$
$\Rightarrow \mathrm{y}=\frac{1}{\mathrm{k}} \mathrm{x}$
$\Rightarrow$ y = Cx where C = $\frac{1}{k}$
View full question & answer→Question 21 Mark
The general solution of the differential equation $e^x dy + (y e^x + 2x) dx = 0$ is
AnswerIt is given that $e^xdy + (ye^x + 2x) dx = 0$
$\Rightarrow \mathrm{e}^{\mathrm{x}} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{ye}^{\mathrm{x}}+2 \mathrm{x}=0$
$\Rightarrow \frac{d y}{d x}+y=-2 x e^{-x}$
This is equation in the form of $\frac{d y}{d x}+p y=Q$ $($where, $p = 1$ and $Q = -2xe^{-x})$
Now, I.F = $\mathrm{e}^{\int \mathrm{pdx}}=\mathrm{e}^{\int \mathrm{dx}}=\mathrm{e}^{\mathrm{x}}$
Thus, the solution of the given differential equation is given by the relation:
$y(I . F .)=\int(Q \times I . F .) d x+C$
$\Rightarrow \mathrm{ye}^{\mathrm{x}}=\int\left(-2 \mathrm{xe}^{-\mathrm{x}} \cdot \mathrm{e}^{\mathrm{x}}\right) \mathrm{dx}+\mathrm{C}$
$\Rightarrow \mathrm{ye}^{\mathrm{x}}=-\int 2 \mathrm{xdx}+\mathrm{C}$
$\Rightarrow$ $ye^x = -x^2 + C$
$\Rightarrow$ $ye^x + x^2 = C$
View full question & answer→Question 31 Mark
The general solution of a differential equation of the type $\frac{d x}{d y}+\mathrm{P}_{1} x=\mathrm{Q}_{1}$ is
AnswerThe integrating factor of the given differential equation
$\frac{\mathrm{d} \mathrm{x}}{\mathrm{dy}}+\mathrm{P}_{1} \mathrm{x}=\mathrm{Q}_{1} \text { is } \mathrm{e}^{\int \mathrm{P}_{1} \mathrm{d} \mathrm{y}}$
Thus, the general solution of the differential equation is given by,
$\mathrm{x}(\mathrm{I.F.})=\int(\mathrm{Q_1} \times \mathrm{I.F.}) \mathrm{dy}+\mathrm{C}$
$\Rightarrow \mathrm{x} . \mathrm{e}^{\int \mathrm{P}_{1} \mathrm{d} \mathrm{y}}=\int\left(\mathrm{Q}_{1} \mathrm{e}^{\int \mathrm{P}_{1} \mathrm{d} \mathrm{y}}\right) \mathrm{d} \mathrm{y}+\mathrm{C}$
View full question & answer→Question 41 Mark
Find the order and degree (if defined) of the differential equation $\frac{d^{4} y}{d x^{4}}-\sin \left(\frac{d^{3} y}{d x^{3}}\right)=0$
AnswerIt is given that the equation is $\frac{d^{4} y}{d x^{4}}-\sin \left(\frac{d^{3} y}{d x^{3}}\right)=0$
We can see that the highest order derivative present in the differential is $\frac{d^{4} y}{d x^{4}}$.
Thus, its order is four.
The given differential equation is not a polynomial equation.
Therefore, its degree is not defined.
View full question & answer→Question 51 Mark
Find the order and degree (if defined) of the differential equation $\left(\frac{d y}{d x}\right)^{3}-4\left(\frac{d y}{d x}\right)^{2}+7 y=\sin x$
AnswerIt is given that equation is $\left(\frac{d y}{d x}\right)^{3}-4\left(\frac{d y}{d x}\right)^{2}+7 y=\sin x$
$\left(\frac{d y}{d x}\right)^{3}-4\left(\frac{d y}{d x}\right)^{2}+7 y-\sin x=0$
We can see that the highest order derivative present in the differential is $\frac{dy}{dx}$.
Thus, its order is one. It is polynomial equation in $\frac{dy}{dx}$. The highest power raised to $\frac{dy}{dx}$ is 3.
Therefore, its degree is three.
View full question & answer→Question 61 Mark
Find the order and degree (if defined) of the differential equation $\frac{d^{2} y}{d x^{2}}+5 x\left(\frac{d y}{d x}\right)^{2}-6 y=\log x$
AnswerIt is given that equation is $\frac{d^{2} y}{d x^{2}}+5 x\left(\frac{d y}{d x}\right)^{2}-6 y=\log x$
$\Rightarrow\frac{d^{2} y}{d x^{2}}+5 x\left(\frac{d y}{d x}\right)^{2}-6 y-\log x=0$
We can see that the highest order derivative present in the differential is $\frac{d^{2} y}{d x^{2}}$.
Thus, its order is two.
The highest power raised to $\frac{d^{2} y}{d x^{2}}$ is 1.
Therefore, its degree is one.
View full question & answer→Question 71 Mark
The Integrating Factor of the differential equation $\left(1-y^{2}\right) \frac{d x}{d y}+y x=a y~(-1<y<1)$ is
AnswerIt is given that $\left(1-y^{2}\right) \frac{d x}{d y}+y x=a y$
$\Rightarrow \frac{d x}{d y}+\frac{y x}{1-y^{2}}=\frac{a y}{1-y^{2}}$
This is equation in the form of $\frac{d x}{d y}+p x=Q$ (where $p=\frac{y}{1-y^{2}}$ and Q = $\frac{ay}{1-y^{2}}$)
Now, I.F. = $\mathrm{e}^{\int \mathrm{p} \mathrm{d} \mathrm{y}}=\mathrm{e}^{\int \frac{\mathrm{y}}{1-\mathrm{y}^{2}} \mathrm{d} \mathrm{y}}=\mathrm{e}^{-\frac{1}{2} \log \left(1-\mathrm{y}^{2}\right)}=\mathrm{e}^{\log \left[\frac{1}{\sqrt{\left(1-\mathrm{y}^{2}\right)}}\right]}$
= $\frac{1}{\sqrt{\left(1-y^{2}\right)}}$
View full question & answer→Question 81 Mark
The Integrating Factor of the differential equation $x \frac{d y}{d x}-y=2 x^{2}$ is
AnswerIt is given that $x \frac{d y}{d x}-y=2 x^{2}$
$\Rightarrow \frac{d y}{d x}-\frac{y}{x}=2 x$
This is equation in the form of $\frac{d y}{d x}+p y=Q ($where, $p=-\frac{1}{x}$ and $Q = 2x^2)$
Now, I.F. = $e^{\int p d x}=e^{\int \frac{-1}{x} d x}=e^{\log \left(x^{-1}\right)}=x^{-1}=\frac{1}{x}$
View full question & answer→Question 91 Mark
Which of the following is a homogeneous differential equation?
Answerit is a homogeneous differential equation ,because the degree of each individual term is same i.e. 2.
View full question & answer→Question 101 Mark
A homogeneous equation of the form $\frac{{dy}}{{dx}} = h\left( {\frac{x}{y}} \right)$ can be solved by making the substitution
AnswerA homogeneous equation of the form $\frac{{dy}}{{dx}} = h\left( {\frac{x}{y}} \right)$ can be solved by making the substitution x= vy.so that it becomes variable separable form and integration is then possible
View full question & answer→Question 111 Mark
The general solution of the differential equation $\frac{d y}{d x}=e^{x+y}$ is
AnswerWe have, $ \frac{d y}{d x}=e^{x+y}$
$\Rightarrow \frac{d y}{d x}=e^{x} \times e^{y}$
separating variables
$\Rightarrow \mathrm{e}^{-\mathrm{y}} \mathrm{dy}=\mathrm{e}^{\mathrm{x}} \mathrm{d} \mathrm{x}$
Integrating both sides
$\Rightarrow \int \mathrm{e}^{-\mathrm{y}} \mathrm{d} \mathrm{y}=\int \mathrm{e}^{\mathrm{x}} \mathrm{d} \mathrm{x}$
$\Rightarrow-e^{-y}=e^{x}+c$
$\Rightarrow e^{x}+e^{-y}=-c$
Or,
$e^{x}+e^{-y}=c$ (c is a constant)
View full question & answer→Question 121 Mark
The number of arbitrary constants in the particular solution of a differential equation of third order are:
Answer0, because the particular solution is free from arbitrary constants.
View full question & answer→Question 131 Mark
The number of arbitrary constants in the general solution of a differential equation of fourth order are:
Answer4, because the no. of arbitrary constants is equal to order of the differential equation.
View full question & answer→Question 141 Mark
Determine order and degree (if defined) of differential equation: $y" + (y')^2 + 2y = 0$
AnswerIt is given that equation is $y’’ + (y’)^2 + 2y = 0$
We can see that the highest order derivative present in the differential is $y”.$
Thus, its order is two.
Highest power raised to $y”$ is $1$
Therefore, its degree is one.
View full question & answer→Question 151 Mark
Determine order and degree (if defined) of differential equation: $y' + y = e^x$
AnswerIt is given that equation is $y’ + y = e^x$
$\Rightarrow y’ + y – e^x = 0$
We can see that the highest order derivative present in the differential is $y’.$
Thus, its order is one. It is polynomial equation in $y’$
So, the highest power raised to y’ is $1$
Therefore, its degree is one.
View full question & answer→Question 161 Mark
Determine order and degree (if defined) of differential equation y'" + 2y" + y' = 0
AnswerIt is given that equation is y″′ +2y” + y’ = 0
We can see that the highest order derivative present in the differential is y”’.
Thus, its order is three. It is polynomial equation in y”’, y” and y’
So, the highest power raised to y”’ is 1
Therefore, its degree is one.
View full question & answer→Question 171 Mark
Determine order and degree (if defined) of differential equation $\left(y^{\prime \prime \prime}\right)^{2}+\left(y^{\prime \prime}\right)^{3}+\left(y^{\prime}\right)^{4}+y^{5}=0$
AnswerIt is given that equation is $(y”’)^2 + (y”)^3 +(y’)^4 + y^5 = 0$
We can see that the highest order derivative present in the differential is $y”’.$
Thus, its order is three.
It is polynomial equation in $y”’, y”$ and $y’$
So, the highest power raised to $y”’$ is $2$
Therefore, its degree is two.
View full question & answer→Question 181 Mark
Determine order and degree (if defined) of differential equation $\frac{d^{2} y}{d x^{2}}$ = cos 3x + sin 3x
AnswerThe given differential equation is $\frac{d^{2} y}{d x^{2}}$ = cos 3x + sin 3x
$\Rightarrow \frac{d^{2} y}{d x^{2}}$ - cos 3x - sin 3x = 0
We can see that the highest order derivative present in the given differential equation is $\frac{d^{2} y}{d x^{2}}$
Thus, its order is two.
It is polynomial equation in $\frac{d^{2} s}{d t^{2}}$ and the power is 1
Therefore, its degree is one.
View full question & answer→Question 191 Mark
Determine order and degree (if defined) of differential equation $\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\cos \left(\frac{d y}{d x}\right)=0$
AnswerIt is given that equation is $\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\cos \left(\frac{d y}{d x}\right)=0$
We can see that the highest order derivative present in the given differential equation is $\frac{d^{2} y}{d x^{2}}$
Thus, its order is two. The given differential equation is not a polynomial equation in its derivative.
Therefore, its degree is not defined.
View full question & answer→Question 201 Mark
Determine order and degree (if defined) of differential equation ${\left( {\frac{{ds}}{{dt}}} \right)^4} + 3s\frac{{{d^2}s}}{{d{t^2}}} = 0$
View full question & answer→Question 211 Mark
Determine order and degree (if defined) of differential equation y' + 5y = 0
AnswerIt is given that equation is y’ + 5y = 0
We can see that the highest order derivative present in the differential is y’.
Thus, its order is one. It is polynomial equation in y’. The highest power raised to y’ is 1.
Therefore, its degree is one.
View full question & answer→Question 221 Mark
The order of the differential equation $2 x^{2} \frac{d^{2} y}{d x^{2}}-3 \frac{d y}{d x}+y=0$ is
AnswerIt is given that equation is $2 x^{2} \frac{d^{2} y}{d x^{2}}-3 \frac{d y}{d x}+y=0$
We can see that the highest order derivative present in the given differential equation is $\frac{d^{2} y}{d x^{2}}$
Thus, its order is two.
View full question & answer→Question 231 Mark
The degree of the differential equation $\left(\frac{d^{2} y}{d x^{2}}\right)^{3}+\left(\frac{d y}{d x}\right)^{2}+\sin \left(\frac{d y}{d x}\right)+1=0$ is
AnswerIt is given that equation is $\left(\frac{d^{2} y}{d x^{2}}\right)^{3}+\left(\frac{d y}{d x}\right)^{2}+\sin \left(\frac{d y}{d x}\right)+1=0$
The given differential equation is not a polynomial equation in its derivative
Therefore, its degree is not defined.
View full question & answer→Question 241 Mark
Determine order and degree (if defined) of differential equation: y" + 2y' + siny = 0
AnswerIt is given that equation is y’’ + 2y’ + siny = 0
We can see that the highest order derivative present in the differential is y”.
Thus, its order is two. It is polynomial equation in y” and y’
So, the highest power raised to y” is 1
Therefore, its degree is one.
View full question & answer→Question 251 Mark
Determine order and degree (if defined) of differential equation: $\frac{{{d^4}y}}{{d{x^4}}} + \sin \left( {y'''} \right)$ = 0
Answerorder = 4, degree = not defined
View full question & answer→Question 261 Mark
Find the order and degree, if defined, of the differential equation $y^{\prime \prime \prime}+y^{2}+e^{y^{\prime}}=0$
AnswerThe highest order derivative present in the differential equation is y′′′, so its order is three. The given differential equation is not a polynomial equation in its derivatives and so its degree is not defined.
View full question & answer→Question 271 Mark
Find the order and degree, if defined, of the differential equation $x y \frac{d^{2} y}{d x^{2}}+x\left(\frac{d y}{d x}\right)^{2}-y \frac{d y}{d x}=0$
AnswerThe highest order derivative present in the given differential equation is $\frac{d^{2} y}{d x^{2}}$, so its order is two. It is a polynomial equation in $\frac{d^{2} y}{d x^{2}}$ and $\frac{d y}{d x}$ and the highest power raised to $\frac{d^{2} y}{d x^{2}}$ is one, so its degree is one.
View full question & answer→Question 281 Mark
Find the order and degree, if defined, of the differential equation $\frac{d y}{d x}-\cos x=0$
AnswerThe highest order derivative present in the differential equation is $\frac{d y}{d x}$, so its order is one. It is a polynomial equation in $\frac{d y}{d x}$ and the highest power raised to $\frac{d y}{d x}$ is one, so its degree is one.
View full question & answer→