MCQ 11 Mark
The number of arbitrary constants in the particular solution of a differential equation of second order is $($are$):$
AnswerIn the particular solution of a differential equation of third order, there is no arbitrary constant because in the particular solution of any differential equation, we remove all the arbitrary constant by substituting some particular values.
View full question & answer→MCQ 21 Mark
The solution of the differential equation $dy = (1 + y^2) dx$ is :
- A
$\text{y}=\tan\text{x}+\text{c}$
- ✓
$\text{y}=\tan(\text{x}+\text{c})$
- C
$\tan^{-1}(\text{y}+\text{c})=\text{x}$
- D
$(\tan^{-1}(\text{y}+\text{c})=2\text{x}$
AnswerCorrect option: B. $\text{y}=\tan(\text{x}+\text{c})$
Given : $dy = (1 + y^2) dx$
$\Rightarrow\frac{\text{dy}}{1+\text{y}^2}\text{dx}$
Integrating both sides, we get
$\Rightarrow\int\frac{\text{dy}}{1+\text{y}^2}=\int\text{dx}$
$\Rightarrow\tan^{-1}\text{y}=\text{x}+\text{c}$
$\Rightarrow\text{y}=\tan(\text{x}+\text{c})$
View full question & answer→MCQ 31 Mark
The degree and the order of the differential equation $\text{y}=\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+\Big(\frac{\text{dx}}{\text{dy}}\Big)^2$ are respectively:
AnswerCorrect option: C. $4, 1$
Given differential equation is
$\text{y}=\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+\Big(\frac{\text{dx}}{\text{dy}}\Big)^2$
$\Rightarrow\text{y}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)^4+1$
Here, degree is $44$ and order is $1.$
View full question & answer→MCQ 41 Mark
The solution of the differential equation $\text{x}\frac{\text{dy}}{\text{dx}}=\text{y}+\text{x}\ \tan\frac{\text{y}}{\text{x}}$ is :
- ✓
$\sin\frac{\text{x}}{\text{y}}=\text{x}+\text{C}$
- B
$\sin\frac{\text{y}}{\text{x}}=\text{Cx}$
- C
$\sin\frac{\text{x}}{\text{y}}=\text{Cy}$
- D
$\sin\frac{\text{y}}{\text{x}}=\text{Cy}$
AnswerCorrect option: A. $\sin\frac{\text{x}}{\text{y}}=\text{x}+\text{C}$
We have,
$\text{x}\frac{\text{dy}}{\text{dx}}=\text{y}+\text{x}\ \tan\frac{\text{y}}{\text{x}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}+\tan\frac{\text{y}}{\text{x}}\ ...(\text{i})$
Let $\text{y}=\upsilon\text{x}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\upsilon+\text{x}\frac{\text{d}\upsilon}{\text{dx}}$
Putting both value in $(i)$
$\upsilon+\text{x}\frac{\text{d}\upsilon}{\text{dx}}=\upsilon+\tan\upsilon$
$\Rightarrow \frac{\text{d}\upsilon}{\tan\upsilon}=\frac{\text{dx}}{\text{x}}$
Integrating both sides, we get
$\log\sin\upsilon=\log\text{x}+\log\text{C}$
$\Rightarrow \log\frac{\sin\upsilon}{\text{x}}=\log\text{C}$
$\Rightarrow \frac{\sin\upsilon}{\text{x}}=\text{C}$
$\Rightarrow\sin\upsilon=\text{Cx}$
$\Rightarrow\sin(\frac{\text{y}}{\text{x}})=\text{Cx}$
View full question & answer→MCQ 51 Mark
The degree of the differntial equation $\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)^{2}=\Big(\frac{\text{dy}}{\text{dx}}\Big)=\text{y}^{3}$ is:
- A
$\frac{1}{2}$
- ✓
$2$
- C
$3$
- D
$4$
AnswerWe have,
$\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)^{2}=\Big(\frac{\text{dy}}{\text{dx}}\Big)=\text{y}^{3}$
The highest order derivative is $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}$ and its power is $2.$
Hence, the degree is $2.$
View full question & answer→MCQ 61 Mark
The solution of differential equation $(\text{e}^\text{y}+1)\cos\text{dx}+\text{e}^\text{y}\sin\text{x}\text{dy}=0$ is:
- ✓
$\text{e}^\text{y}+1\sin\text{x}=\text{c}$
- B
$\text{e}^\text{y}\sin=\text{c}$
- C
$(\text{e}^\text{y}+1)\cos\text{x}=\text{c}$
- D
AnswerCorrect option: A. $\text{e}^\text{y}+1\sin\text{x}=\text{c}$
View full question & answer→MCQ 71 Mark
The solution of the differential equation $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{1+\text{x}^2}$ is:
- A
$\text{y}=\frac{1}{2}\log|2+\text{x}^2|+\text{c}$
- B
$\text{y}=\frac{1}{2}\log(1+\text{x})+\text{c}$
- ✓
$\text{y}=\log\Big(\sqrt{1+\text{x}^2}\Big)+\text{c}$
- D
AnswerCorrect option: C. $\text{y}=\log\Big(\sqrt{1+\text{x}^2}\Big)+\text{c}$
View full question & answer→MCQ 81 Mark
Which of the following differentials equation has $y = x$ as one of its particular solution?
- A
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{x}^{2}\frac{\text{dy}}{\text{dx}}+\text{xy}=\text{x}$
- B
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{x}^{2}\frac{\text{dy}}{\text{dx}}+\text{xy}=\text{x}$
- ✓
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{x}^{2}\frac{\text{dy}}{\text{dx}}+\text{xy}=\text{0}$
- D
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{x}^{2}\frac{\text{dy}}{\text{dx}}+\text{xy}=\text{0}$
AnswerCorrect option: C. $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{x}^{2}\frac{\text{dy}}{\text{dx}}+\text{xy}=\text{0}$
We have,
$\text{y}=\text{x}\ ...(\text{i})$
Differentiating both sides of $(i)$ with respect to $x$, we get
$\frac{\text{dy}}{\text{dx}}=1\ ...(\text{ii})$
Differentiating both sides of $(ii)$ with respect to $x,$ we get
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=0$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{x}^{2}=\text{x}^{2}$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{x}\times\text{x}=\text{x}^{2}\times1$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{xy}=\text{x}^{2}\times1$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{xy}=\text{x}^{2}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{x}^{2}\frac{\text{dy}}{\text{dx}}+\text{xy}=0$
View full question & answer→MCQ 91 Mark
The differential equation $\frac{\text{dx}}{\text{dx}}=\frac{1}{\text{ax}+\text{by}+\text{c}},$ where $a, b, c$ are all non zero real numbers, is:
AnswerCorrect option: B. Linear in $x$
$=\frac{\text{dx}}{\text{dx}}={\text{ax}+\text{by}+\text{c}}$
$=\frac{\text{dx}}{\text{dx}}-{\text{ax}=\text{by}+\text{c}}$
$=$Linear in $x$
View full question & answer→MCQ 101 Mark
The solution of the differential equartion $\frac{\text{dy}}{\text{dx}}-\frac{\text{y}(\text{x}+1)}{\text{x}}=0$ is given by :
- ✓
$\text{y}=\text{xe}^{\text{x}+\text{C}}$
- B
$\text{x}=\text{ye}^{\text{x}}$
- C
$\text{y}=\text{x}+\text{c}$
- D
$\text{xy}=\text{e}^{\text{x}}+\text{C}$
AnswerCorrect option: A. $\text{y}=\text{xe}^{\text{x}+\text{C}}$
We have,
$\frac{\text{dy}}{\text{dx}}-\frac{\text{y}(\text{x}+1)}{\text{x}}=0$
$\Rightarrow \frac{\text{dy}}{\text{dx}}-\frac{\text{y}(\text{x}+1)}{\text{x}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{(\text{x}+1)}{\text{x}}\text{dx}$
Integrating both sides, we get
$ \int\frac{\text{dy}}{\text{y}}=\int\frac{(\text{x}+1)}{\text{x}}\text{dx}$
$ \Rightarrow \int\frac{\text{dy}}{\text{y}}=\int\text{dx}+\int\frac{1}{\text{x}}\text{dx}$
$ \Rightarrow \log{\text{y}}=\text{x}+\log\text{x}+\text{C}$
$\Rightarrow \log{\text{y}}-\log\text{x}=\text{x}+\text{C}$
$ \Rightarrow \log\frac{{\text{y}}}{\text{x}}=\text{x}+\text{C}$
$\Rightarrow \frac{\text{y}}{\text{x}}=\text{e}^{\text{x}+\text{C}}$
$\Rightarrow{\text{y}}=\text{xe}^{\text{x}+\text{C}}$
View full question & answer→MCQ 111 Mark
What is the order of differential equation $y\ " + 5y\ ’ + 6 = 0\ ?$
AnswerGiven, differential equation $y\ ’’ + 5y\ ’ + 6 = 0.$
The highest order derivative present in the differential equation is $y\ ’’$.
Hence, the order is $2.$
View full question & answer→MCQ 121 Mark
The Solution of $\cos(\text{x}+\text{y})\text{ dy}=\text{dx}$ is:
- ✓
$\text{y}=\tan\Big(\frac{\text{x}+\text{y}}{2}\Big)+\text{c}$
- B
$\text{y}=\cos^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)+\text{c}$
- C
$\text{y}=\text{x}\sec\Big(\frac{\text{y}}{\text{x}}\Big)+\text{c}$
- D
AnswerCorrect option: A. $\text{y}=\tan\Big(\frac{\text{x}+\text{y}}{2}\Big)+\text{c}$
View full question & answer→MCQ 131 Mark
Choose the correct answer from the given four option. The degree of the differential equation $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\Big(\frac{\text{d}\text{y}}{\text{d}\text{x}}\Big)^3+6\text{y}^5=0$ is :
Answer$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\Big(\frac{\text{d}\text{y}}{\text{d}\text{x}}\Big)^3+6\text{y}^5=0$
We know that, the degree of a differential equation is highest exponent of order dervivatibve.
$\therefore\text{Degree}=1$
View full question & answer→MCQ 141 Mark
A homogeneous dofferential equation of the from $\frac{\text{dx}}{\text{dy}}=\text{h}(\frac{\text{x}}{\text{y}})$ can be solved by making the substitution :
- A
$y = vx$
- B
$v = yx$
- ✓
$x = vy$
- D
$x = v$
AnswerCorrect option: C. $x = vy$
A homogeneous differential of the from $\frac{\text{dx}}{\text{dy}}=\text{h}(\frac{\text{x}}{\text{y}})$ can be solved by sunstituting $x = vy.$
View full question & answer→MCQ 151 Mark
The solution of the differential equation $2\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=3$ resresents :
AnswerWe have,
$2\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=3$
$\Rightarrow 2\text{x}\frac{\text{dy}}{\text{dx}}=3+\text{y}$
$\Rightarrow \frac{1}{3+\text{y}}\text{dy}=\frac{1}{2\text{x}}\text{dx}$
Interating both sides, we get
$\Rightarrow \int\frac{1}{3+\text{y}}\text{dy}=\frac{1}{2}\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow \log|3+\text{y}|=\frac{1}{2}\log|\text{x}|+\log|\text{C}|$
$\Rightarrow\log|\frac{3+\text{y}}{\sqrt{\text{x}}}|=\log\text{C}$
$\Rightarrow\frac{3+\text{y}}{\sqrt{\text{x}}}=\text{C}$
$\Rightarrow 3+\text{y}=\text{C}\sqrt{\text{x}}$
Squaring both sides, we get
$(3+\text{y})^{2}=\text{C}{\text{x}}\ ...(\text{i})$
Thus, $(i)$ the equation of parabolas.
View full question & answer→MCQ 161 Mark
Choose the correct answer from the given four option. Solution of differential equation $x\ dy - y \ dx = 0$ represents :
- A
- B
Parabola whose vertex is at origin.
- ✓
Straight line passing through origin.
- D
A circle whose centre is at origin.
AnswerCorrect option: C. Straight line passing through origin.
Given that, $\text{x dy}-\text{y dx}=0$
$\Rightarrow\text{x dy}=\text{y dx}$
$\Rightarrow\frac{\text{dy}}{\text{y}}=\frac{\text{dx}}{\text{x}}$
On integrating both sides, we get
$\Rightarrow\int\frac{\text{dy}}{\text{y}}=\int\frac{\text{dx}}{\text{x}}$
$\Rightarrow\log\text{y}=\log\text{x}+\log\text{C}$
$\Rightarrow\log\text{y}=\log\text{Cx}$
$\Rightarrow\text{y}=\text{Cx}$
Which is a straight line passing through origin.
View full question & answer→MCQ 171 Mark
The general solution of differention eqution of the $e^x d y+\left(y e^x+2 x\right) d x=0$ is :
- A
$\mathrm{xe}^y+\mathrm{x}^2=C$
- B
$\mathrm{xe}^y+\mathrm{y}^2=c$
- ✓
$\mathrm{ye}^{\mathrm{x}}+\mathrm{y}^2=C$
- D
$\mathrm{ye}^{\mathrm{y}}+\mathrm{x}^2=c$
AnswerCorrect option: C. $\mathrm{ye}^{\mathrm{x}}+\mathrm{y}^2=C$
We have,
$e^x d y+\left(y e^x+2 x\right) d x=0$
Diving both sides by we get,
$\frac{\text{dy}}{\text{dx}}+(\text{y}+\frac{2\text{x}}{\text{e}^{x}})=0$
$\Rightarrow \frac{\text{dy}}{\text{dx}}+\text{y}=-\frac{2\text{x}}{\text{e}^{\text{x}}}$
Comping with $\frac{\text{dy}}{\text{dx}}=\text{Q}$ we get,
$\text{P}=1, \text{Q}=-\frac{2\text{x}}{\text{e}^{\text{x}}}$
Now,
$\text{I.F}=\text{e}^{\int\text{dx}}$
$=\text{e}^{\text{x}}$
Solution is given by,
$\text{y}\times\text{I.F}=\int(\text{Q}\times\text{I.F}) \text{dx}+\text{C}$
$\Rightarrow \text{ye}^{\text{x}}=-\int\text{e}^{\text{x}}\times \frac{2\text{x}}{\text{e}^{\text{x}}}\text{dx}+\text{C}$
$\Rightarrow \text{ye}^{\text{x}}=-2\int\text{x}\ \text{dx}+\text{C}$
$\Rightarrow \text{ye}^{\text{x}}=-\text{x}^{2}+\text{C}$
$\Rightarrow \text{ye}^{\text{x}}+\text{x}^{2}=\text{C}$
View full question & answer→MCQ 181 Mark
The order of differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}-\Big(\frac{\text{dy}}{\text{dx}^2}\Big)=1$ is :
AnswerThe order of differential equation is the order of thehighest derivative in the equation
$\therefore$ the above given equation is of second order
View full question & answer→MCQ 191 Mark
The solution of the differential equation $\frac{\text{dy}}{\text{dx}}-\text{Ky}=0, \text{y}(0)=1$ approaches to zero when $\text{x}\rightarrow\propto$ if,
- A
$K = 0$
- B
$K > 0$
- ✓
$K < 0$
- D
AnswerCorrect option: C. $K < 0$
We have,
$\Rightarrow \frac{\text{dy}}{\text{dx}}-\text{Ky}=0$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\text{Ky}$
$\Rightarrow \frac{1}{\text{y}}\text{dy}=\text{K}\ \text{dx}$
Integrating both sides, we get
$ \int\frac{1}{\text{y}}\text{dy}=\text{K}\int\text{dx}$
$\Rightarrow \log|\text{y}|=\text{Kx}+\text{C}\ ...(\text{i})$
Now,
$\text{y}(0)=1$
$\text{C}=0$
Putting $C = 0$ in $(i),$
$\log|\text{y}|=\text{Kx}$
$\Rightarrow \text{e}^{\text{Kx}}=\text{y}$
According to the quation,
$\text{e}^{\text{K}\propto}=0$
View full question & answer→MCQ 201 Mark
The solution of $\frac{\text{dy}}{\text{dx}}=1+\text{x}+\text{y}+\text{xy}$ is:
- A
$\text{x}-\text{y}=\text{k}(1+\text{xy})$
- ✓
$\log(1+\text{y})=\text{x}+\frac{\text{x}^2}{2}+\text{k}$
- C
$\log(1+\text{y})=\text{x}+\frac{\text{y}^2}{2}=\text{k}$
- D
AnswerCorrect option: B. $\log(1+\text{y})=\text{x}+\frac{\text{x}^2}{2}+\text{k}$
View full question & answer→MCQ 211 Mark
A homogeneous differential equation of the from $\frac{\text{dx}}{\text{dy}}=\text{h}\Big(\frac{\text{x}}{\text{y}}\Big)$ can be solved by making the substitution.
- A
$y = vx$
- B
$v = yx$
- ✓
$x = vy$
- D
$x = v$
AnswerCorrect option: C. $x = vy$
We know that a homogeneous differential equation of the form $\frac{\text{dx}}{\text{dy}}=\text{h}\Big(\frac{\text{x}}{\text{y}}\Big)$ can be solved by the substitution $\frac{\text{x}}{\text{y}}=\text{v}$ i.e., ${x}=\text{vy}.$
View full question & answer→MCQ 221 Mark
Choose the correct answer from the given four options.The general solution of the differential equation $(\text{e}^{\text{x}}+1)\text{ydy}=(\text{y}+1)\text{e}^{\text{x}}$ is :
- A
$(\text{y}+1)=\text{k}(\text{e}^{\text{x}}+1)$
- B
$\text{y}+1=\text{e}^{\text{x}}+1+\text{k}$
- ✓
$\text{y}=\log\left\{\text{k}(\text{y}+1)(\text{e}^{\text{x}}+1)\right\}$
- D
$\text{y}=\log\left\{\frac{\text{e}^{\text{x}}+1}{\text{y}+1}\right\}+\text{k}$
AnswerCorrect option: C. $\text{y}=\log\left\{\text{k}(\text{y}+1)(\text{e}^{\text{x}}+1)\right\}$
Given differential equation
$(\text{e}^{\text{x}}+1)\text{ydy}=(\text{y}+1)\text{e}^{\text{x}}\text{dx}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^{\text{x}}(1+\text{y})}{(\text{e}^{\text{x}}+1)\text{y}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{(\text{e}^{\text{x}}+1)\text{y}}{\text{e}^{\text{x}}(1+\text{y})}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{e}^{\text{x}}\text{y}}{\text{e}^{\text{x}}(1+\text{y})}\frac{\text{y}}{\text{e}^{\text{x}}(1+\text{y})}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{1+\text{y}}+\frac{\text{y}}{(1+\text{y})\text{e}^{\text{x}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{1+\text{y}}\Big(1+\frac{1}{\text{e}^{\text{x}}}\Big)\text{dx}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{1+\text{y}}\Big(1+\frac{1}{\text{e}^{\text{x}}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{1+\text{y}}\Big(\frac{\text{e}^{\text{x}}+1}{\text{e}^{\text{x}}}\Big)$
$\Rightarrow\Big(\frac{\text{y}}{1+\text{y}}\Big)\text{dy}=\Big(\frac{\text{e}^{\text{x}}}{\text{e}^{\text{x}}+1}\Big)\text{dx}$
On integrating both sides, we get
$\int\frac{\text{y}}{1+\text{y}}\text{dy}=\int\frac{\text{e}^{\text{x}}}{1+\text{e}^{\text{x}}}\text{dx}$
$\Rightarrow\int\frac{1+\text{y}-1}{1+\text{y}}\text{dy}=\int\frac{\text{e}^{\text{x}}}{1+\text{e}^{\text{x}}}\text{dx}$
$\Rightarrow\int1\text{dy}-\int\frac{\text{y}}{1+\text{y}}\text{dy}=\int\frac{\text{e}^{\text{x}}}{1+\text{e}^{\text{x}}}\text{dx}$
$\Rightarrow\text{y}-\log|(1+\text{y})+\log(1+\text{e}^{\text{x}})|+\log(\text{k})$
$\Rightarrow\text{y}=\log\left\{\text{k}(1+\text{y})(1+\text{e}^{\text{x}})\right\}$
View full question & answer→MCQ 231 Mark
The order of the differential equation $\Bigg[1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^5\Bigg]^\frac{2}{3}=\frac{\text{d}^3\text{y}}{\text{dx}}$ is :
- A
$2$
- B
$1$
- ✓
$3$
- D
$\frac{2}{3}$
Answer$=\Bigg[1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^5\Bigg]^\frac{2}{3}=\frac{\text{d}^3\text{y}}{\text{dx}}$
$\Rightarrow\Bigg[1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^5\Bigg]^2=\Big(\frac{\text{d}^3\text{y}}{\text{dx}}\Big)^3$
Hence order of above differential equation is $3$
View full question & answer→MCQ 241 Mark
Which of the following is a homogeneous differential equation?
- A
$ (4 x+6 y+5) d y-(3 y+2 x+4) d x=0 $
- B
$(x y) d x-\left(x^3+y^3\right) d y=0 $
- C
$ \left(x^3+2 y^2\right) d x+2 x y\ \ d y=0 $
- ✓
$ y^2 d x+\left(x^2-x y-y^2\right) d y=0 $
AnswerCorrect option: D. $ y^2 d x+\left(x^2-x y-y^2\right) d y=0 $
Out of the given four options, option $(D)$ is the only option in which all coefficients of $dx$ and $dy$ are of same degree i.e., $2$. It may be noted that $xy$ is a term of second degree.
Hence differential equation in option $(D)$ is Homogeneous differential equation.
View full question & answer→MCQ 251 Mark
The number of arbitrary constants in the general solution of a differential equation of fourth order are:
AnswerThe number of arbitrary constants in a solution of a differential equation of order nn is equal to its order.
So, here it is $4.$
View full question & answer→MCQ 261 Mark
The differential equation satisfied by $\text{ax}^{2}+\text{by}^{2}=1$ is :
- A
$\text{xyy}_{2}+\text{y}_{1}^{2}+\text{yy}_{1}=0$
- ✓
$\text{xyy}_{2}+\text{xy}_{1}^{2}-\text{yy}_{1}=0$
- C
$\text{xyy}_{2}+\text{xy}_{1}^{2}+\text{yy}_{1}=0$
- D
AnswerCorrect option: B. $\text{xyy}_{2}+\text{xy}_{1}^{2}-\text{yy}_{1}=0$
We have,
$\text{ax}^{2}+\text{by}^{2}=1\ ...(\text{i})$
Differential both sides of $(i)$ with $x,$ we get
$2\text{ax}+2\text{by}\frac{\text{dy}}{\text{dy}}=0\ ...(\text{ii})$
Differential both sides of $(ii)$ with $x,$ we get
$2\text{ax}+2\text{b}\Big(\frac{\text{dy}}{\text{dy}}\Big)^{2}+2\text{}by\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=0$
$\Big[\text{y}\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big(\frac{\text{dy}}{\text{dx}^{2}}\Big)\Big]=-\frac{2\text{a}}{2\text{b}}$
$\text{x}\Big[\text{y}\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big(\frac{\text{dy}}{\text{dx}^{2}}\Big)\Big]=-\Big(-\frac{\text{y}}{\text{x}}\frac{\text{dy}}{\text{dx}}\Big)$
$\text{xy}\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{x}\Big(\frac{\text{dy}}{\text{dx}}\big)^{2}-\text{y}\frac{\text{dy}}{\text{dx}}=0$
$\text{xyy}_{2}+\text{x}(\text{y}_{1}^{2})-\text{yy}_{1}=0$
View full question & answer→MCQ 271 Mark
Choose the correct answer from the given four option. The solution of the differential equation $\cos\text{x}\ \sin\text{y}\ \text{dx}+\sin\text{x}\ \cos\text{y}\ \text{dy}=0$ is :
- A
$\frac{\\sin\text{x}}{\sin\text{y}}=\text{C}$
- ✓
$\sin\text{x}\ \sin\text{y}=\text{C}$
- C
$\sin\text{x}+\sin\text{y}=\text{C}$
- D
$\cos\text{x}\ \cos\text{y}=\text{C}$
AnswerCorrect option: B. $\sin\text{x}\ \sin\text{y}=\text{C}$
Given differential equation is
$\cos\text{x}\ \sin\text{y}\ \text{dx}+\sin\text{x}\ \cos\text{y}\ \text{dy}=0$
$\Rightarrow\cos\text{x}\ \sin\text{y}\ \text{dx}=-\sin\text{x}\ \cos\text{y}\ \text{dy}=0$
$\Rightarrow\frac{\cos\text{x}}{\sin\text{x}}\text{dx}=-\frac{\cos\text{y}}{\sin\text{y}}\text{dy}$
$\Rightarrow\cot\text{x}\ {\text{dx}}=-\cot\text{y}\ {\text{dy}}$
On integrating both sides, we get
$\log\ \sin\text{x}=-\log\ \sin\text{y}+\log\ \text{C}$
$\Rightarrow\log\ \sin\text{x}\ \sin\text{y}=\log\ \text{C}$
Carrying the exponent on both sides, we get
$\Rightarrow\sin\text{x}\ \sin\text{y}=\text{C}$
View full question & answer→MCQ 281 Mark
The degree of the differential equation $2\text{x}^{2}\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+3\frac{\text{dy}}{\text{dx}}+\text{y}=0$ is :
AnswerWe have,
$2\text{x}^{2}\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+3\frac{\text{dy}}{\text{dx}}+\text{y}=0$
Here, the highest order is $\frac{\text{d}^{2}\text{y}}{\text{d}^{2}\text{x}}.$
Hence, the order is $2.$
View full question & answer→MCQ 291 Mark
$\text{y}=\sin kt$ satisfies the differential equation $y\ "+\ 9y = 0$. Then $k$ :
- ✓
$\pm3$
- B
$0$
- C
$\pm2$
- D
$\pm4$
AnswerCorrect option: A. $\pm3$
Given, $\text{y}=\sin kt$ satisfies the differential equation $y\ " + 9y = 0.$
Then, we have,$-\text{k}^2\sin\text{kt}+9\sin\text{kt}=0$ or,
$=\text{k}^2-9=0$
$[$Since $\sin\text{kt}\neq 0] $ or,
$=\text{k}=\pm3.$
View full question & answer→MCQ 301 Mark
The integrating factor of the differential equation $(\text{x}\log\text{x})\frac{\text{dy}}{\text{dx}}+\text{y}=2\ \log\text{x}$ is given by :
- A
$\log(\log\text{x})$
- B
$\text{e}^{\text{x}}$
- ✓
$\log\text{x}$
- D
$\text{x}$
AnswerCorrect option: C. $\log\text{x}$
We have,
$(\text{x}\log\text{x})\frac{\text{dy}}{\text{dx}}+\text{y}=2\ \log\text{x}$
Dividing both sides by,
$\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}+\log\text{x}}=\frac{2}{\text{x}}$
$\frac{\text{dy}}{\text{dx}}+\Big(\frac{\text{1}}{\text{x}+\log\text{x}}\Big)\text{y}=\frac{2}{\text{x}}$
Comparing with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
$\text{P}=\frac{1}{\text{x}\log\text{x}}$
$\text{Q}=\frac{2}{\text{x}}$
Now,
$\text{I.F}=\text{e}^{\int\text{P}\text{dx}}$
$\text{I.F}=\text{e}^{\int\frac{1}{\text{x}\log\text{x}}}\text{dx}$
$=\text{e}^{\log(\log\text{x})}$
$=\log\text{x}$
View full question & answer→MCQ 311 Mark
Choose the correct answer from the given four options.The solution of the differential equation $\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}-\text{y}}+\text{x}^2\text{e}^{-\text{y}}$ is :
- A
$\text{y}=\text{e}^{\text{x}-\text{y}}-\text{x}^2\text{e}^{-\text{y}}+\text{c}$
- ✓
$\text{e}^{\text{y}}-\text{e}^{\text{x}}=\frac{\text{x}^3}{3}+\text{c}$
- C
$\text{e}^{\text{x}}+\text{e}^{\text{y}}=\frac{\text{x}^3}{3}+\text{c}$
- D
$\text{e}^{\text{x}}-\text{e}^{\text{y}}=\frac{\text{x}^3}{3}+\text{c}$
AnswerCorrect option: B. $\text{e}^{\text{y}}-\text{e}^{\text{x}}=\frac{\text{x}^3}{3}+\text{c}$
We have, $\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}-\text{y}}+\text{x}^2\text{e}^{-\text{y}}$
$\Rightarrow\text{e}^{\text{y}}\text{dy}=(\text{e}^{\text{x}}+\text{x}^2)\text{dx}$
$\Rightarrow\int\text{e}^{\text{y}}\text{dy}=\int(\text{e}^{\text{x}}+\text{x}^2)\text{dx}$
$\Rightarrow\text{e}^{\text{y}}=\text{e}^{\text{x}}+\frac{\text{x}^3}{3}+\text{C}$
$\Rightarrow\text{e}^{\text{y}}-\text{e}^{\text{x}}=\frac{\text{x}^3}{3}+\text{C}$
View full question & answer→MCQ 321 Mark
What are the order and degree, respectively, of the differential equation : $\Big(\frac{\text{d}^3\text{y}}{\text{dx}^3}\Big)^2=\text{y}^4+\Big(\frac{\text{dy}}{\text{dx}}\Big)^5\ ?$
AnswerCorrect option: C. $3, 2$
Order is the highest derivative of the dependent variable with respect to the independent variable and degree is the highest power to which the highest order derivative in the differential equation is raised.
so, Order $= 3$ and Degree $=2.$
View full question & answer→MCQ 331 Mark
Choose the correct answer from the given four options.The curve for which the slope of the tangent at any point is equal to the ratio of the abcissa to the ordinate of the point is :
AnswerSlope of langent to the curve $=\frac{\text{dy}}{\text{dx}}$
According to the question, $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{\text{y}}$
$\Rightarrow\text{y dy}=\text{x dx}$
On integrating both sides, we get
$\frac{\text{y}^2}{2}=\frac{\text{x}^2}{2}+\text{C}$
$\Rightarrow\text{y}^2-\text{x}^2=2\text{C}$ which is an equation of rectangular hyperbola.
View full question & answer→MCQ 341 Mark
The solution of the differention $\frac{\text{dy}}{\text{dx}}+1=\text{e}^{\text{x}+\text{y}}$ is :
- A
$(\text{x}+\text{y})\text{e}^{\text{x}+\text{y}}=0$
- B
$(\text{x}+\text{C})\text{e}^{\text{x}+\text{y}}=0$
- C
$(\text{x}-\text{C})\text{e}^{\text{x}+\text{y}}=1$
- ✓
$(\text{x}-\text{C})\text{e}^{\text{x}+\text{y}}+1=0$
AnswerCorrect option: D. $(\text{x}-\text{C})\text{e}^{\text{x}+\text{y}}+1=0$
We have,
$\frac{\text{dy}}{\text{dx}}+1=\text{e}^{\text{x}+\text{y}}$
Let $\text{x}+\text{y}=\text{u}$
$\Rightarrow 1+\frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}+1=\frac{\text{du}}{\text{dx}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\text{e}^{\text{u}}$
$\Rightarrow \text{e}^{-\text{u}}\text{du}=\text{dx}$
Intergrating both sides, we get
$\Rightarrow \text{e}^{-\text{u}}=\text{x}-\text{C}$
$\Rightarrow -1=\text{e}^{-\text{u}}(\text{x}-\text{C})$
$\Rightarrow (\text{x}-\text{C})\text{e}^{\text{x}+\text{y}}+1=0$
View full question & answer→MCQ 351 Mark
What is the degree of the differential equation : $\text{y}=\text{x}\frac{\text{dy}}{\text{dx}}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^{-2}\ ?$
AnswerOrder : The order of a differential equation is the order of the highest derivative appearing in it.
Degree : The degree of a differential equation is the power of the highest derivative occurring in it, after the Equation has been expressed in a form free from radicals as far as the derivatives are concerned.
Given :
$\text{y}=\text{x}\frac{\text{dy}}{\text{dx}}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^{-2}\text{y} $
$=\text{x}\frac{\text{dy}}{\text{dx}}+\frac{1}{\Big(\frac{\text{dy}}{\text{dx}}\Big)2}\text{y}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$
$=\text{x}\frac{\text{dy}}{\text{dx}}^3+1$
For the given differential equation the highest order derivative is $1.$
Now, the power of the highest order derivative is $3.$
We know that the degree of a differential equation is the power of the highest derivative.
Hence, the degree of the differential equation is $3.$
View full question & answer→MCQ 361 Mark
If $\left(x^2+y^2\right) d y=xy \ dx, y(1)=1$, and $y\left(x_0\right)=e$, then $x_0=$
AnswerCorrect option: C. $\sqrt{3}\text.{e}$
View full question & answer→MCQ 371 Mark
Integrating factor of the differntial equation $\cos\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\sin\text{x}=1$ is :
- A
$\sin\text{x}$
- ✓
$\sec\text{x}$
- C
$\tan\text{x}$
- D
$\cos\text{x}$
AnswerCorrect option: B. $\sec\text{x}$
We have,
$\cos\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\sin\text{x}=1$
Dividing both sides by we get
$\frac{\text{dy}}{\text{dx}}+\frac{\sin\text{x}}{\cos\text{x}}\text{y}=\frac{1}{\cos\text{x}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}+(\tan\text{x})\text{y}=\frac{1}{\cos\text{x}}$
Comparing with $ \frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
$\text{P}=\tan\text{x}$
$\text{Q}=\frac{1}{\cos\text{x}}$
Now,
$\text{I.F}=\text{e}^{\int\tan\text{x}}\text{dx}$
$=\text{e}^{\log(\sec\text{x})}$
$=\sec\text{x}$
View full question & answer→MCQ 381 Mark
Choose the correct answer from the given four options. The solution of $\frac{\text{dy}}{\text{dx}}+\text{y}=\text{e}^{-\text{x}}, y(0) = 0$ is :
- A
$\text{y}=\text{e}^{-\text{x}}(\text{x}-1)$
- B
$\text{y}=\text{x}\text{e}^{\text{x}}$
- C
$\text{y}=\text{x}\text{e}^{-\text{x}}+1$
- ✓
$\text{y}=\text{x}\text{e}^{-\text{x}}$
AnswerCorrect option: D. $\text{y}=\text{x}\text{e}^{-\text{x}}$
We have, $\frac{\text{dy}}{\text{dx}}+\text{y}=\text{e}^{-\text{x}}$
Which is a linear differential equation.
Here, $\text{P}=1$ and $\text{Q}=\text{e}^{-\text{x}}$
$\text{I.F.}=\text{e}^{\int\text{dx}}=\text{e}^{\text{x}}$
The general solution is
$\text{y}.\text{e}^{-\text{x}}=\int\text{e}^{-\text{x}}.\text{e}^{\text{x}}\ \text{dx}+\text{C}$
$\Rightarrow\text{y}\text{e}^{\text{x}}=\int\text{dx}+\text{C}$
$\Rightarrow\text{y}\text{e}^{-\text{x}}=\text{x}+\text{C}\ ....(\text{i})$
Given, when $x = 0$ and $y = 0$
$\Rightarrow0=0+\text{C}$
$\Rightarrow\text{C}=0$
Eq, $(i)$ resuces to $\text{y}.\text{e}^{-\text{x}}=\text{x}$ or $\text{y}=\text{x}\text{e}^{-\text{x}}.$
View full question & answer→MCQ 391 Mark
Choose the correct answer from the given four options. Family $\text{y}=\text{A}\text{x}+\text{A}^3$ of curves will correspond to a differential equation of order:
AnswerGiven family of curves is $\text{y}=\text{A}\text{x}+\text{A}^3\ .....(\text{i})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{A}$
Replacting $A$ by $\frac{\text{dy}}{\text{dx}}$ in Eq. $(i),$ we get
$\text{y}=\text{x}\frac{\text{dy}}{\text{dx}}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^3$
$\therefore$ Order $=1$
View full question & answer→MCQ 401 Mark
If $\frac{\text{dy}}{\text{dx}}=\text{e}^{-2\text{y}}$ and $y = 0,$ when $x = 5,$ then the value of $x$ when $y = 3$ is:
- A
$\text{e}^5$
- B
$\text{e}^5+1$
- ✓
$\frac{\text{e}^5+9}{2}$
- D
$\log_\text{e}6$
AnswerCorrect option: C. $\frac{\text{e}^5+9}{2}$
View full question & answer→MCQ 411 Mark
Determine the degree and order of the given differential equation respectively : $\text{y}=\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big(\frac{\text{dx}}{\text{dy}}\Big)^{-2}\ ?$
- A
$1, 2$
- B
$2, 1$
- C
$1, 4$
- ✓
$4, 1$
AnswerCorrect option: D. $4, 1$
Order : The order of a differential equation is the order of the highest derivative appearing in it.
Degree : The degree of a differential equation is the power of the highest derivative occurring in it, after the Equation has been expressed in a form free from radicals as far as the derivatives are concerned.
Note :
Degree is defined if function is a polynomial, if differential contains logarithmic, exponential and trigonometric function of the highest derivative, then degree is not defined.
Degree and order is always a positive integer.
Given : $\text{y}=\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big(\frac{\text{dx}}{\text{dy}}\Big)^{-2}$
To find : Order Degree
$\text{y}=\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\frac{1}{\Big(\frac{\text{dy}}{\text{dx}}\Big)^2}$
$\text{y}=\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=\text{x}\Big(\frac{\text{dx}}{\text{dy}}\Big)^4+1$
Hence, the degree is $4$ order is $1$.
View full question & answer→MCQ 421 Mark
Find the degree of the differential equation : $\Big(1+\frac{\text{dx}}{\text{dy}}\Big)^3=\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$
AnswerGiven, the differential equation is :
$\Big(1+\frac{\text{dx}}{\text{dy}}\Big)^3=\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$
We can expand it and get :
$1+3\frac{\text{dx}}{\text{dy}}^3+3\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+\Big(\frac{\text{dy}}{\text{dx}}\Big)^3=\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$
The exponent of highest derivative is the degree.
Therefore, the degree is $3.$
View full question & answer→MCQ 431 Mark
The differential equation which respresents the famliy of curves $\text{y}=\text{e}^{\text{Cx}}$ is :
- A
$\text{y}_{1}=\text{C}^{2}\text{y}$
- B
$\text{xy}_{1}-\log\text{y}=0$
- C
$\text{x}\log\text{y}=\text{yy}_{1}$
- ✓
$\text{y}\log\text{y}=\text{xy}_{1}$
AnswerCorrect option: D. $\text{y}\log\text{y}=\text{xy}_{1}$
We have,
$\text{y}=\text{e}^{\text{Cx}}$
Taking in both sides, we get
$\Rightarrow \log\ \text{y}=\text{Cx}\ ...(\text{1})$
Differentiating both sides of $(i)$ with respect to $x,$ we get
$\frac{1}{\text{y}_{1}}=\text{C}$
Substituting the value of $C $ in in $(i)$. we get
$\log\ \text{y}=\frac{\text{y}_{1}}{\text{y}}\text{x}$
$\Rightarrow \text{y}\ \log\ \text{y}=\text{y}_{1}\text{x}$
View full question & answer→MCQ 441 Mark
Order of $\Big( \frac{\text{dy}}{\text{dx}}\Big)^{3}+ \Big( \frac{\text{dy}}{\text{dx}}\Big)^{2}+\text{y}^{4}=0$ is :
Answer$=\Big( \frac{\text{dy}}{\text{dx}}\Big)^{3}+ \Big( \frac{\text{dy}}{\text{dx}}\Big)^{2}+\text{y}^{4}=0$
$=\text{i.e} {\text{ y}_{1}}^{3.}+{\text{y}_{1}}^{2}+\text{y}^{4}=0$
$=$ Order $= 1$
$=$ Degree $= 3$
View full question & answer→MCQ 451 Mark
The equation of the curve whose slope is given by $\frac{\text{dy}}{\text{dx}}=\frac{2\text{y}}{\text{x}};\text{x}>0,\text{y} > 0$ and which passes through the point $(1, 1)$ is :
- ✓
$\text{x}^{2}=\text{y}$
- B
$\text{y}^{2}=\text{x}$
- C
$\text{x}^{2}=2\text{y}$
- D
$\text{y}^{2}=2\text{x}$
AnswerCorrect option: A. $\text{x}^{2}=\text{y}$
We have,
$\frac{\text{dy}}{\text{dx}}=\frac{2\text{y}}{\text{x}}$
$\Rightarrow\frac{1}{2}\times\frac{1}{\text{y}}\text{dy}=\frac{1}{\text{x}}\text{dx}$
Interating both sides, we get
$\Rightarrow\frac{1}{2}\int\frac{1}{\text{y}}\text{dy}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\frac{1}{2}\ \log{\text{y}}=\log{\text{x}}+\log\text{C}$
$\Rightarrow\log{\text{y}}^{\frac{1}{2}}-\log{\text{x}}=\log\text{C}$
$\Rightarrow\log\big(\frac{\sqrt{\text{y}}}{2}\big)=\log\text{C}$
$\Rightarrow\frac{\sqrt{\text{y}}}{2}=\text{C}$
$\Rightarrow\sqrt{\text{y}}=\text{Cx}\ ...(\text{i})$
As $(i)$ passes through $(1, 1),$ we get
$1=\text{C}$
Putting the value of $C$ in $(1),$ we get
$\Rightarrow\sqrt{\text{y}}=\text{x}$
$\Rightarrow{\text{y}}=\text{x}^{2}$
View full question & answer→MCQ 461 Mark
The order and degree of the differential equation $\left(y^{\prime \prime \prime}\right)^2+\left(y^{\prime \prime}\right)^3-\left(y^{\prime}\right)^4+y^5=0$ is :
- ✓
$3$ and $2$
- B
$1$ and $2$
- C
$2$ and $3$
- D
$1$ and $4$
AnswerCorrect option: A. $3$ and $2$
The given differential equation is $\left(y^{\prime \prime \prime}\right)^2+\left(y^{\prime \prime}\right) 2-\left(y^{\prime}\right)^3+y^5=0$
Clearly, its order is $3$ and degree is $2$.
Hence, option $3$ and $2$ is correct
View full question & answer→MCQ 471 Mark
The solution of the differential equation, $\text{x}^2\frac{\text{dy}}{\text{dx}}.\cos\frac{1}{\text{x}}-\text{y}\sin\frac{1}{\text{x}}=-1,$ where $\text{y}\rightarrow-1$ as $\text{x}\rightarrow-\infty,$ is :
- ✓
$\text{y}=\sin\frac{1}{\text{x}}-\cos\frac{1}{\text{x}}$
- B
$\text{y}=\frac{\text{x}+1}{\text{x}\sin\frac{1}{\text{x}}}$
- C
$\text{y}=\cos\frac{1}{\text{x}}+\sin\frac{1}{\text{x}}$
- D
$\text{y}=\frac{\text{x}+1}{\text{x}\cos\frac{1}{\text{x}}}$
AnswerCorrect option: A. $\text{y}=\sin\frac{1}{\text{x}}-\cos\frac{1}{\text{x}}$
View full question & answer→MCQ 481 Mark
Choose the correct answer from the given four option. $\text{y}= \text{a}\text{e}^{\text{mx}}+\text{b}\text{e}^{-\text {mx}}$ satisfies which of the following differential equation?
- A
$\frac{\text{dy}}{\text{dx}}+\text {my}=0$
- B
$\frac{\text{dy}}{\text{dx}}-\text {my}=0$
- ✓
$\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{m} ^2\text{y}=0$
- D
$\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{m} ^2\text{y}=0$
AnswerCorrect option: C. $\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{m} ^2\text{y}=0$
Given that, $\text{y}=\text{a}\text{e}^{\text{mx}}+\text{b}\text{e}^{- \text{mx}}$
On differentiating both sides $\text{w.r.t. x},$ we get
$\frac{\text{dy}}{\text{dx}}=\text {ma}\text{e}^{\text{mx}}+\text{bm}\text{e}^{-\text {mx}}$
Again, differentiating both sides $\text{w.r.t. x},$ we get
$\frac{\text{d}^2\text{y}}{\text {d}\text{x}^2}=\text{m}^2\text{a}\text{e}^{\text{mx}}+ \text{b}\text{m}^2\text{e}^{-\text{mx}}$
$\Rightarrow\frac{\text{d}^2\text {y}}{\text{d}\text{x}^2}=\text{m}^2(\text{a}\text{e}^{\text {mx}}+\text{b}\text{e}^{-\text{mx}})$
$\Rightarrow\frac{\text{d}^2\text {y}}{\text{d}\text{x}^2}=\text{m}^2\text{y}$
$\Rightarrow\frac{\text{d}^2\text {y}}{\text{d}\text{x}^2}-\text{m}^2\text{y}=0$
View full question & answer→MCQ 491 Mark
Choose the correct answer from the given four option. $\tan^{-1}+\tan^{-1}\text{y}=\text{C}$ is the general solution of the differential equation :
- A
$\frac{\text{d}\text{y}}{\text{d}\text{x}}=\frac{1+\text{y}^2}{1+\text{x}^2}$
- B
$\frac{\text{d}\text{y}}{\text{d}\text{x}}=\frac{1+\text{x}^2}{1+\text{y}^2}$
- ✓
$(1+\text{x}^2)\text{dy}+(1+\text{y}^2)\text{dx}=0$
- D
$(1+\text{x}^2)\text{dx}+(1+\text{y}^2)\text{dy}=0$
AnswerCorrect option: C. $(1+\text{x}^2)\text{dy}+(1+\text{y}^2)\text{dx}=0$
Given is, $\tan^{-1}+\tan^{-1}\text{y}=\text{C}$
On differentiating above eqaution $\text{w.r. t. x,}$ we get
$\frac{1}{1+\text{x}^2}+\frac{1}{1+\text{y}^2}.\frac{\text{d}\text{y}}{\text{d}\text{x}}=0$
$\Rightarrow\frac{1}{1+\text{y}^2}.\frac{\text{d}\text{y}}{\text{d}\text{x}}=-\frac{1}{1+\text{x}^2}$
$\Rightarrow(1+\text{x}^2)\text{dy}+(1+\text{y}^2)\text{dx}=0$
View full question & answer→MCQ 501 Mark
The solution of $\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}=\frac{1}{\sqrt{1+\text{x}^2}}$ is:
- A
$\text{y}=\frac{1+\text{x}^2}{\text{x}}+\frac{\text{c}}{\text{x}}$
- ✓
$\text{y}=\frac{\sqrt{1+\text{x}^2}}{\text{x}}+\frac{\text{c}}{\text{x}}$
- C
$\text{y}=\frac{\text{x}}{\sqrt{1+\text{x}^2}}+\text{cx}$
- D
AnswerCorrect option: B. $\text{y}=\frac{\sqrt{1+\text{x}^2}}{\text{x}}+\frac{\text{c}}{\text{x}}$
View full question & answer→