Question 13 Marks
Form the differential equation corresponding to $\text{y}^2=\text{a}(\text{b}-\text{x}^2)$ bt eliminating a and b.
Answer$\text{y}^2=\text{a}(\text{b}-\text{x}^2)$
Differential it with respect to x,
$2\text{y}\frac{\text{dy}}{\text{dx}}=\text{a}(-2\text{x}) ...(1)$
Again, differential it with respect to x,
$2\Big[\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\frac{\text{dy}}{\text{dx}}\times\frac{\text{dy}}{\text{dx}}\Big]=-2\text{a}$
$\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=-\Big(\frac{2\text{y}}{-2\text{x}}\frac{\text{dy}}{\text{dx}}\Big)$
using equation (1)
$\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=\frac{\text{y}}{\text{x}}\frac{\text{dy}}{\text{dx}}$
$\text{x}\Big\{\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big\}=\text{y}\frac{\text{dy}}{\text{dx}}$
View full question & answer→Question 23 Marks
Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\frac{1-\cos2\text{y}}{1+\cos2\text{y}}$
AnswerWe have
$\frac{\text{dy}}{\text{dx}}=\frac{1-\cos2\text{y}}{1+\cos2\text{y}}$
$=\frac{2\sin^2\text{y}}{2\cos^2\text{y}}$
$\frac{\text{dy}}{\text{dx}}=\tan^2\text{y}$
$\frac{\text{dy}}{\tan^2\text{y}}=\text{dx}$
$\int\cot^2\text{y dy}=\int\text{dx}$
$\int(\text{cosec}^2\text{y}-1)\text{dy}=\int\text{dx}$
$-\cot\text{y}-\text{y}+\text{C}=\text{x}$
$\text{C}=\text{x}+\text{y}+\cot\text{y}$
View full question & answer→Question 33 Marks
Solve the following differential equations:
$\frac{\text{dy}}{\text{dx}}=2\text{xy, y}(0)=1$
Answer$\frac{\text{dy}}{\text{dx}}=2\text{xy, y}(0)=1$
$\int\frac{\text{dy}}{\text{y}}=\int2\text{x dx}$
$\log|\text{y}|=2\frac{\text{x}^2}{2}+\text{C}$
$\log|\text{y}|=\text{x}^2+\text{C}...(1)$
Put $\text{x}=0,\text{y}=1$
$\log(1)=0+\text{C}$
$0=0+\text{C}$
$\text{C}=0$
Put $\text{C}=0$ in equation (1),
$\log\text{y = x}^2$
$\text{y = e}^{\text{x}^{2}}$
View full question & answer→Question 43 Marks
Find the equation of the curve which passes through the origin and has the slope x + 3y − 1 at any point (x, y) on it.
AnswerAccording to the quation,
$\frac{\text{dy}}{\text{dx}}=\text{x}+3\text{y}-1$
$\Rightarrow \frac{\text{dy}}{\text{dx}}-3\text{y}=\text{x}-1$
Comparing with we get,
$\text{P}=-3, \text{Q}=\text{x}-1$
Now,
$\text{I.F}=\text{e}^-{\int3\text{dx}}$
$=\text{e}^{-3\text{x}}$
So, the solution is given by
$\text{y}\times\text{I.F}=\int\text{Q}\times\text{I.F}\ \text{dx}+\text{C}$
$\Rightarrow \text{ye}^{-3\text{x}}=\int(\text{x}-1)\text{e}^{-3\text{x}}\ \text{dx}+\text{C}$
$\Rightarrow \text{ye}^{-3\text{x}}=\int\text{x}\text{e}^{-3\text{x}}\ \text{dx}-\int\text{e}^{-3\text{x}}\ \text{dx}+\text{C}$
$\Rightarrow \text{ye}^{-3\text{x}}=\text{x}\int\text{e}^{-3\text{x}}\ \text{dx}-\int\big[ \frac{\text{d}}{\text{dx}}(\text{x})\int\text{e}^{-3\text{x}}\text{dx}\big]\text{dx}-\text{e}^{-3\text{dx}}+\text{C}$
$\Rightarrow \text{ye}^{-3\text{x}}=\text{x}\int\text{e}^{-3\text{x}}\ \text{dx}+\frac{1}{3} \int\text{e}^{-3\text{x}}\text{dx}-\int\text{e}^{-3\text{x}}\text{dx}+\text{C}$
$\Rightarrow \text{ye}^{-3\text{x}}=-\frac{1}{3}\text{xe}^{-3\text{x}}+\frac{1}{3}\int\text{e}^{-3\text{x}}\text{dx}+\text{C}$
$\Rightarrow \text{y} =-\frac{1}{3}\text{x}+\frac{1}{9}+\frac{1}{3}+\text{Ce}^{-3\text{x}}$
$\Rightarrow \text{y} =-\frac{1}{3}\text{x}+\frac{2}{9}+\text{Ce}^{-3\text{x}}$
Since the curve passes throught this origin it equation,
$\Rightarrow 0=-0+\frac{2}{9}+\text{Ce}^{0}$
$\Rightarrow \text{C}=-\frac{2}{9}$
Putting the value of C in the equation of the curve, we get
$\Rightarrow \text{y} =-\frac{1}{3}\text{x}+\frac{2}{9}(1-\text{e}^{-3\text{x}})$
$\Rightarrow \text{y} +\frac{1}{3}\text{x}=\frac{2}{9}(1-\text{e}^{-3\text{x}})$
$\Rightarrow 3(3\text{y}+\text{x})=2(1-\text{e}^{-3\text{x}})$
View full question & answer→Question 53 Marks
Solve the following differential equations:$\frac{\text{dy}}{\text{dx}}+\frac{\cos\text{x}\sin\text{y}}{\cos\text{y}}=0$
Answer$\frac{\text{dy}}{\text{dx}}+\frac{\cos\text{x}\sin\text{y}}{\cos\text{y}}=0$
$\frac{\text{dy}}{\text{dx}}=-\cos\text{x}\tan\text{y}$
$\frac{\text{dy}}{\tan\text{y}}=-\cos\text{x dx}$
$\int\cot\text{ y dy}=-\int\cos\text{x dx}$
$\log|\sin\text{y}|=-\sin\text{x + C}$
$\sin\text{x}+\log|\sin\text{y}|=\text{C}$
View full question & answer→Question 63 Marks
For the following differntial equations verify that the accompanying function is a solution:
| Differential equation |
Function |
| $\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}=0$ |
$\text{y}=\pm\sqrt{\text{a}^2-\text{x}^2}$ |
AnswerWe have
$\text{y}=\pm\sqrt{\text{a}^2-\text{x}^2}$
$\Rightarrow\text{y}^2=\text{a}^2-\text{x}^2\ ...(1)$
Given differential equation $\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}=0$
Differentiating both sides of (1) with respect to x, we get
$2\text{y}\frac{\text{dy}}{\text{dx}}=-2\text{x}$
$\Rightarrow\text{y}\frac{\text{dy}}{\text{dx}}={\text{x}}$
$\Rightarrow\text{x}+\text{y}\frac{\text{dy}}{\text{dx}}=0$
Hence, the given function is the solution to the given differential equation.
View full question & answer→Question 73 Marks
Verify that $\text{y}=4\sin3\text{x}$ is a solution of the differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}+9\text{y}=0.$
AnswerWe have,
$\text{y}=4\sin3\text{x}\ ...(1)$
Differentiating both sides of equation (1) with respect to 3, we get
$\frac{\text{dy}}{\text{dx}}=12\cos3\text{x}\ ...(2)$
Differentiating both sides of equation (2) with respect to 3, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-36\sin3\text{x}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-9(4\sin3\text{x})$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-9\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}+9\text{y}=0$
Hence, the given function is the solution to the given differential equation.
View full question & answer→Question 83 Marks
Solve the following differential equations:$\frac{\text{dr}}{\text{dt}}=-\text{rt, r}(0)=\text{r}_{0}$
Answer$\frac{\text{dr}}{\text{dt}}=-\text{rt, r}(0)=\text{r}_{0}$
$\int\frac{\text{dr}}{\text{r}}=-\int\text{tdt}$
$\log|\text{r}|=-\frac{\text{t}^2}{2}+\text{C}...(1)$
Put $\text{t = 0, r = r}_{0}$ inequation (1),
$\log|\text{r}_0|=0+\text{C}$
$\log|\text{r}_0|=\text{C}$
Now,
$\log|\text{r}|=-\frac{\text{t}^2}{2}+\log|\text{r}_0|$
$\frac{\text{r}}{\text{r}_0}=\text{e}^{-\frac{\text{t}^2}{2}}$
$\text{r}=\text{r}_0\text{e}^{-\frac{\text{t}^2}{2}}$
View full question & answer→Question 93 Marks
Solve the following differential equations:
$\frac{\text{dy}}{\text{dx}}=(1+\text{x}^2)(1+\text{y}^2)$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}=(1+\text{x}^2)(1+\text{y}^2)$
$\Rightarrow\frac{1}{(1+\text{y}^2)}\text{dy}=(1+\text{x}^2)\text{dx}$
Integrating both sides, we get
$\int\frac{1}{(1+\text{y}^2)}\text{dy}=\int(1+\text{x}^2)\text{dx}$
$\Rightarrow\tan^{-1}\text{y = x}+\frac{\text{x}^3}{3}+\text{C}$
Hence, $\tan^{-1}\text{y = x}+\frac{\text{x}^3}{3}+\text{C}$ is the required solution.
View full question & answer→Question 103 Marks
Represent the following families of curves by forming the corresponding differential equation:
$(\text{x}-\text{a})^2-\text{y}^2=1$
AnswerThe equation of the family of curves is
$(\text{x}-\text{a})^2-\text{y}^2=1\ ...(1)$
where a is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
$2(\text{x}-\text{a})-2\text{y}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow(\text{x}-\text{a})-\text{y}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\sqrt{1+\text{y}^2}=\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow1+\text{y}^2=\text{y}^2\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$
$\Rightarrow\text{y}^2\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-\text{y}^2=1$
It is the required differential equation.
View full question & answer→Question 113 Marks
Solve the following differential equation
$\text{x}\frac{\text{dy}}{\text{dx}}+\cot\text{y}=0$
AnswerWe have,$\text{x}\frac{\text{dy}}{\text{dx}}+\cot\text{y}=0$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}-\cot\text{y}$
$\Rightarrow\frac{1}{\text{x}}\ \text{dx}=-\frac{1}{\cot\text{y}}\ \text{dy}$
$\Rightarrow\frac{1}{\text{x}}\ \text{dx}=-\tan\text{y dy}$
Integrating both sides, we get
$\int\frac{1}{\text{x}}\ \text{dx}=-\int\tan\text{y dy}$
$\Rightarrow\int|\text{x}|=-\int|\sec\text{y}|+\int\text{C}$
$\Rightarrow\int|\text{x}|=-\int|\cos\text{y}|+\int\text{C}$
$\Rightarrow\text{x}=\text{C}\cos\text{y}$
Hence, $\text{x}=\text{C}\cos\text{y}$ is the required solution.
View full question & answer→Question 123 Marks
Solve the following differential equations:$2\text{x}\frac{\text{dy}}{\text{dx}}=5\text{y},\text{y}(1)=1$
Answer$2\text{x}\frac{\text{dy}}{\text{dx}}=5\text{y},\text{y}(1)=1$
$\int\frac{2\text{dy}}{\text{y}}=\int\frac{5\text{dx}}{\text{x}}$
$2\log|\text{y}|=5\log|\text{x}|+\text{C}...(1)$
Put $\text{x}=1,\text{y}=1$
$2\log(1)=5\log(1)+\text{C}$
$0=\text{C}$
put $\text{C}=0$ in equation (1),
$2\log|\text{y}|=5\log|\text{x}|$
$\text{y}^2=|\text{x}|^5$
$\text{y}=|\text{x}|^{\frac{5}{2}}$
View full question & answer→Question 133 Marks
Solve the following differential equations:$2\text{x}\frac{\text{dy}}{\text{dx}}=3\text{y},\text{y}(1)=2$
Answer$2\text{x}\frac{\text{dy}}{\text{dx}}=3\text{y},\text{y}(1)=2$
$\int\frac{2\text{dy}}{\text{y}}=\int\frac{3\text{dx}}{\text{x}}$
$2\log|\text{y}|=3\log|\text{x}|+\log\text{C}$
$\text{y}^2=\text{x}^3\text{C}...(1)$
Put $\text{x}=1,\text{y}=2$
$4=\text{C}$
Put $\text{C}=4$ in equation (1),
$\text{y}^2=4\text{x}^3$
View full question & answer→Question 143 Marks
Solve the following differential equation
$5\frac{\text{dy}}{\text{dx}}=\text{e}^\text{x}\text{y}^4$
AnswerWe have $5\frac{\text{dy}}{\text{dx}}=\text{e}^\text{x}\text{y}^4$ $\Rightarrow\frac{5}{\text{y}^4}\text{dy}=\text{e}^\text{x dx}$Integrating both sides, we get
$\int\frac{5}{\text{y}^4}\ \text{dy}=\int\text{e}^\text{x}\text{dx}$ $\Rightarrow\frac{-5}{3\text{y}^3}=\text{e}^\text{x}+\text{C}$ Hence, $\frac{-5}{3\text{y}^3}=\text{e}^\text{x}+\text{C}$ is the required solution.
View full question & answer→Question 153 Marks
Find the solution of the differential equation $\text{x}\sqrt{1+\text{y}^{2}}\text{dx}+\text{y}\sqrt{1+\text{x}^{2}}\text{dy}=0.$
Answer$\text{x}\sqrt{1+\text{y}^{2}}\text{dx}+\text{y}\sqrt{1+\text{x}^{2}}\text{dy}=0$
$\Rightarrow \text{y}\sqrt{1+\text{x}^{2}}\text{dy}=-\text{x}\sqrt{1+\text{y}^{2}}\text{dx}=0$
$\Rightarrow \frac{\text{y}}{\sqrt{1+\text{y}^{2}}}\text{dy}=\frac{-\text{x}}{\sqrt{1+\text{x}^{2}}}\text{dx}$
$\Rightarrow \int\frac{\text{y}}{\sqrt{1+\text{y}^{2}}}\text{dy}=-\int\frac{\text{x}}{\sqrt{1+\text{x}^{2}}}\text{dx}$
Let $1+\text{y}^{2}=\text{t}^{2}=\text{t}^{2}, 1+\text{x}^{2}=\text{p}^{2}$
$\Rightarrow 2\text{y}\ \text{dy}=2\text{t}\ \text{dt}, 2\text{x}\ \text{dx}=2\text{p}\ \text{dp}$
$\Rightarrow \text{y}\ \text{dy}=\text{t}\ \text{dt}, \text{x}\ \text{dx}=\text{p}\ \text{dp}$
Substituting in above equation, we get
$\Rightarrow \int\text{dt}=-\int\text{dp}$
$\Rightarrow \text{t}=-\text{p}+\text{C}$
$\Rightarrow\sqrt{1+\text{x}^{2}}+\sqrt{1+\text{y}^{2}}=\text{C}$
View full question & answer→Question 163 Marks
Solve the following differential equations:$\text{ye}^{\frac{\text{x}}{\text{y}}}\text{dx}=(\text{xe}^{\frac{\text{x}}{\text{y}}}+\text{y}^2)\text{dy, y}\neq0$
Answer$\text{ye}^{\frac{\text{x}}{\text{y}}}\text{dx}=(\text{xe}^{\frac{\text{x}}{\text{y}}}+\text{y}^2)\text{dy}$
$\Rightarrow\text{ye}^{\frac{\text{x}}{\text{y}}}\text{dx = xe}^{\frac{\text{x}}{\text{y}}}\text{dy}+\text{y}^2\text{dy}$
$\Rightarrow\text{ye}^{\frac{\text{x}}{\text{y}}}\text{dx}-\text{xe}^{\frac{\text{x}}{\text{y}}}\text{dy = y}^2\text{dy}$
$\Rightarrow(\text{ydx}-\text{xdy})\text{e}^{\frac{\text{x}}{\text{y}}}=\text{y}^2\text{dy}$
$\Rightarrow\frac{(\text{ydx}-\text{xdy})}{\text{y}^2}\text{e}^{\frac{\text{x}}{\text{y}}}=\text{dy}$
$\Rightarrow\text{e}^{\frac{\text{x}}{\text{y}}}\text{d}\Big(\frac{\text{x}}{\text{y}}\Big)=\text{dy}$
$\Rightarrow\int\text{e}^{\frac{\text{x}}{\text{y}}}\text{d}\Big(\frac{\text{x}}{\text{y}}\Big)=\int\text{dy}$
$\Rightarrow\text{e}^{\frac{\text{x}}{\text{y}}}=\text{y + C}$
View full question & answer→Question 173 Marks
Show that $\text{y}=\text{ax}^3+\text{bx}^2+\text{c}$ is a solution of the differential equation $\frac{\text{d}^3\text{y}}{\text{dx}^3}=6\text{a}$
AnswerWe have,
$\text{y}=\text{ax}^3+\text{bx}^2+\text{c}\ ...(1)$
Differentiating both sides of (1) with respect in x, we get
$\frac{\text{dy}}{\text{dx}}=3\text{ax}^2+2\text{bx}\ ...(2)$
Differentiating both sides of (2) with respect in x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=6\text{ax}+2\text{b}\ ...(3)$
Differentiating both sides of (3) with respect in x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=6\text{a}$
Hence, the given function is the solution to the given differential equation.
View full question & answer→Question 183 Marks
Represent the following families of curves by forming the corresponding differential equation:
$\text{y}=\text{ax}^3$
AnswerThe equation of the family of curves is
$\text{y}=\text{ax}^3\ ...(1)$
where a is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=3\text{ax}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=3\times\frac{\text{y}}{\text{x}^3}\times\text{x}^2$
$\Rightarrow\text{x}\frac{\text{dy}}{\text{dx}}=3\text{y}$
It is the required differential equation.
View full question & answer→Question 193 Marks
Represent the following families of curves by forming the corresponding differential equation:
$\text{x}^2+(\text{y}-\text{b})^2=1$
AnswerThe equation of the family of curves is
$\text{x}^2+(\text{y}-\text{b})^2=1\ ...(1)$
where a is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
$2\text{x}+2(\text{y}-\text{b})\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow2\text{x}+2\sqrt{1-\text{x}^2}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\text{x}=-\sqrt{1-\text{x}^2}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\text{x}^2=(1-\text{x}^2)\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$
$\Rightarrow\text{x}^2=\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-\text{x}^2\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$
$\Rightarrow\text{x}^2\Big[1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]=\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$
It is the required differential equation.
View full question & answer→Question 203 Marks
Solve the following differential equations:$\frac{\text{dy}}{\text{dx}}=\text{y}\sin2\text{x, y}(0)=1$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}=\text{y}\sin2\text{x, y}(0)=1$
$\Rightarrow\frac{1}{\text{y}}\text{dy}=\sin2\text{x dx}$
Integrating both sides, we get
$\int\frac{1}{\text{y}}\text{dy}=\int\sin2\text{x dx}$
$\Rightarrow\log|\text{y}|=-\frac{\cos 2\text{x}}{2}+\text{C}...(1)$
Given: $\text{x}=0,\text{y}=1.$
Substituting the values of x and y in (1), we get
$\log|1|=-\frac{1}{2}+\text{C}$
$\Rightarrow\text{C}=\frac{1}{2}$
Substituting the values of C in (1), we get
$\log|\text{y}|=-\frac{\cos 2\text{x}}{2}+\frac{1}{2}$
$\Rightarrow\log|\text{y}|=\frac{1-\cos 2\text{x}}{2}$
$\Rightarrow\log|\text{y}|=\sin^2\text{x}$
$\Rightarrow\text{y}=\text{e}^{\sin^2\text{x}}$
Hence, $\text{y}=\text{e}^{\sin^2\text{x}}$ is the required solution
View full question & answer→Question 213 Marks
For the following differntial equations verify that the accompanying function is a solution:
| Differential equation |
Function |
| $\text{x}^3\frac{\text{d}{^2}\text{y}}{\text{dx}^2}=1$ |
$\text{y}=\text{ax}+\text{b}+\frac{1}{2\text{x}}$ |
AnswerWe have
$\text{y}=\text{ax}+\text{b}+\frac{1}{2\text{x}}\ ...(1)$
Differentiating both sides of (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=\text{a}-\frac{1}{2\text{x}^2}\ ...(2)$
Now, differentiating both sides of (2) with respect to x, we get
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\Big(-\frac{1}{2}\Big)\times\Big(\frac{-2}{\text{x}^3}\Big)$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1}{\text{x}^3}$
$\Rightarrow\text{x}^3\frac{\text{d}^2\text{y}}{\text{dx}^2}=1$
Hence, the given function is the solution to the given differential equation.
View full question & answer→Question 223 Marks
verify that $\text{y}=\text{cx}+2\text{c}^2$ is a solution of the differential equation $2\Big(\frac{\text{d}\text{y}}{\text{dx}}\Big)^2-\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}=0$
AnswerWe have,
$\text{y}=\text{cx}+2\text{c}^2\ ...(1)$
Differentiating both sides of (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=\text{c}\ ...(2)$
Now,
$2\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}$
$=2\text{c}^2+\text{cx}-\text{cx}-2\text{c}^2=0$
$\Rightarrow\Big(\frac{\text{d}\text{y}}{\text{dx}}\Big)^2+\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=0$
Hence, the given is the solution to the given differential equation.
View full question & answer→Question 233 Marks
Solve the following differential equations:
$\text{xy}\frac{\text{dy}}{\text{dx}}=1+\text{x + y + xy}$
Answer$\text{xy}\frac{\text{dy}}{\text{dx}}=1+\text{x + y + xy}$
$=(1+\text{x})+\text{y}(1+\text{x})$
$\text{xy}\frac{\text{dy}}{\text{dx}}=(1+\text{x})(1+\text{y})$
$\int\frac{\text{ydy}}{\text{y}+1}=\int\frac{1+\text{x}}{\text{x}}\text{dx}$
$\int\Big(1-\frac{1}{\text{y}+1}\Big)\text{dy}=\int\Big(\frac{1}{\text{x}}+1\Big)\text{dx}$
$\text{y}-\log|\text{y}+1|=\log|\text{x}|+\text{x}+\log|\text{c}|$
$\text{y}=\log|\text{cx(y+1})|+\text{x}$
View full question & answer→Question 243 Marks
Differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{y}=0,\text{y}(0)=0,\text{y}'(0)=1$Function $\text{y}=\sin\text{x}$
AnswerWe have,$\text{y}=\sin{\text{x}} ...(1)$
Differentiating both sides of (1) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=\cos\text{x} ...(2)$
Differentiating both sides of (2) with respect to x, we get
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=-\sin{\text{x}}$
$\Rightarrow\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=-\text{y}$ [Using (1)]
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{y}=0$
it is the given differential equation.
Here, $\text{y}=\sin\text{x}$ satisfies the given differential equation; hence, it is a solution.
Also, when $\text{x}=0,\text{y}=\sin0=0,\text{i.e.},\text{y}(0)=0.$
And, when $\text{x}=0,\text{y}'=\cos 0=1,\text{i.e.,}\text{y}'(0)=1.$
Hence, $\text{y}=\sin\text{x}$ is the solution to the given initial value problem.
View full question & answer→Question 253 Marks
Solve the following differential equations:$(1-\text{x}^2)\text{dy + xy dx = xy}^2\text{ dx}$
View full question & answer→Question 263 Marks
Represent the following families of curves by forming the corresponding differential equation:
$\text{y}=4\text{ax}$
AnswerThe equation of the family of curves is
$\text{y}=4\text{ax}\ ...(1)$
where a is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
$2\text{y}\frac{\text{dt}}{\text{dx}}=4\text{a}$
$\Rightarrow2\text{y}\frac{\text{dy}}{\text{dx}}=\frac{\text{y}^2}{\text{x}}$
$\Rightarrow\text{y}-2\text{x}\frac{\text{dy}}{\text{dx}}=0$
It is the required differential equation.
View full question & answer→Question 273 Marks
show that $\text{y}=\text{be}^\text{x}+\text{ce}^{2\text{x}}$ is a solution of the differential equation, $\frac{\text{d}^2\text{y}}{\text{dx}^2}-3\frac{\text{dy}}{\text{dx}}+2\text{y}=0$
Answer$\text{y}=\text{be}^\text{x}+\text{ce}^{2\text{x}}\ ...(1)$
Differentiating both sides with respect to x,
$\frac{\text{dy}}{\text{dx}}=\text{be}^\text{x}+2\text{ce}^{2\text{x}}\ ...(2)$
Differentiating both sides with respect to x,
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{be}^\text{x}+4\text{ce}^{2\text{x}}\ ...(3)$
now,
$\frac{\text{d}^2\text{y}}{\text{dx}^2}-3\frac{\text{dy}}{\text{dx}}+2\text{y}$
$=\text{be}^\text{x}+4\text{ce}^{2\text{x}}-3(\text{be}^\text{x}+2\text{ce}^{2\text{x}})+2(\text{be}^\text{x}+\text{ce}^{2\text{x}})$
$=\text{be}^\text{x}+4\text{ce}^{2\text{x}}-3\text{be}^\text{x}+6\text{ce}^{2\text{x}}+2\text{be}^\text{x}+2\text{ce}^{2\text{x}}$
$=3\text{be}^\text{x}-3\text{be}^\text{x}+6\text{ce}^{2\text{x}}-6\text{ce}^{2\text{x}}$
$=0$
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Differential equation $\frac{\text{dy}}{\text{dx}}=\text{y},\text{y}(0)=1$Function $\text{y}=\text{e}^\text{x}$
Answerwe have,$y=e^x$ ...(1) Differentiating both sides of (1)with respect to x, we get
$=\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}}$
$=\frac{\text{dy}}{\text{dx}}=\text{y}$ [Using (1)]
It is the given differential equation.
Here, y = ex satisfies the given differential equation; hence, it is a solution.
Also, when $x = 0, y = e^0= 1$, i.e. y(0) = 1
Hence, $y = e^x$ is the solution to the given initial value problem.
View full question & answer→Question 293 Marks
Solve the following differential equations:
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x+y}}+\text{e}^{-\text{x+y}}$
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x+y}}+\text{e}^{-\text{x+y}}$
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x+y}}+\text{e}^{-\text{x}+\text{y}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{y}}(\text{e}^{\text{x}}+\text{e}^{-\text{x}})$
$\Rightarrow\text{e}^{-\text{y}}\text{dy}=(\text{e}^{\text{x}}+\text{e}^{-\text{x}})\text{dx}$
Integrating both sides, we get
$\int\text{e}^{-\text{y}}\text{dy}=\int(\text{e}^{\text{x}}+\text{e}^{-\text{x}})\text{dx}$
$\Rightarrow-\text{e}^{-\text{y}}=\text{e}^{\text{x}}-\text{e}^{-\text{x}}+\text{C}$
$\Rightarrow\text{e}^{-\text{x}}-\text{e}^{-\text{y}}=\text{e}^{\text{x}}+\text{C}$
Hence, $\text{e}^{-\text{x}}-\text{e}^{-\text{y}}=\text{e}^{\text{x}}+\text{C}$ is the required solution.
View full question & answer→Question 303 Marks
Represent the followinf families of curves by forming the corresponding differential equation:
$\text{y}=\text{e}^{\text{ax}}$
AnswerThe equation of the family of curves is
$\text{y}=\text{e}^{\text{ax}}\ ...(1)$
$\Rightarrow\log\text{y}=\text{ax}$
where a is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
$\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\text{a}$
$\Rightarrow\frac{1}{\text{y}}\frac{\text{dy}}{\text{dx}}=\frac{\log\text{y}}{\text{x}}$
$\Rightarrow\text{x}\frac{\text{dx}}{\text{dx}}=\text{y}\log\text{y}$
It is the required differential equation.
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Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\text{x}^5+\text{x}^2-\frac{2}{\text{x}},\text{x}\ne0$
AnswerWe have,$\frac{\text{dy}}{\text{dx}}=\text{x}^5+\text{x}^2-\frac{2}{\text{x}},\text{x}\ne0$
$\Rightarrow\text{dy}=\Big(\text{x}^5+\text{x}^2-\frac{2}{\text{x}}\Big)\text{dx}$
Intergrating both sides, we get
$\Rightarrow\int\text{dy}=\int\Big(\text{x}^5+\text{x}^2-\frac{2}{\text{x}}\Big)\text{dx}$
$\Rightarrow\text{y}=\frac{\text{x}^6}{6}+\frac{\text{x}^3}{3}-2\log|\text{x}|+\text{C}$
Clearly, $\Rightarrow\text{y}=\frac{\text{x}^6}{6}+\frac{\text{x}^3}{3}-2\log|\text{x}|+\text{C}$ is defined for all $\text{x}\in\text{R}$ except x = 0
Hence, $\Rightarrow\text{y}=\frac{\text{x}^6}{6}+\frac{\text{x}^3}{3}-2\log|\text{x}|+\text{C}$, where $\text{x}\in\text{R}-\{0\},$ is the solution o the given differential equation.
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Differential equation $\frac{\text{dy}}{\text{dx}}+\text{y}=0,\text{y}(0)=3$Function $\text{y}=\text{e}^\text{-x}+2$
AnswerHere, $\text{y}=\text{e}^{\text{x}}+1 ....(1)$
Differentiating it with respect to $x,$
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}}$
$\frac{\text{dy}}{\text{dx}}=\text{y}-1 ...(2)$
Again, differentiating it with respect to $x,$
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=\frac{\text{dy}}{\text{dx}}$
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\frac{\text{dy}}{\text{dx}}=0$
It is given differential equation. so,
$y = e^x+ 1$ is a solution of the equation
put $x - 0$ in equation $(1),$
$\Rightarrow y = e^0+ 1 = 2$
$y(0) = 2$
put $x = 0$ in equation $ (2),$
$y' = e^0 = 1$
$y(0) = 1$
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Differential equation $\text{x}\frac{\text{dy}}{\text{dx}}=1,\text{y}(1)=0$Function $\text{y}=\log\text{x}$
AnswerHere, y = logxDifferentiating it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}}$
$\text{x}\frac{\text{dy}}{\text{dx}}=1$
so, y=logx is a solution of the equation
If $\text{x}=1,\text{y}=\log1=0$
so,
y(1) = 0
View full question & answer→Question 343 Marks
Show that $\text{Ax}^2+\text{By}^2=1$ is a solution of the differential equation $\text{x}\Big\{\text{y}=\text{x}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dx}}{\text{dy}}\Big)^2\Big\}=\text{y}\frac{\text{dy}}{\text{dx}}.$
AnswerWe have,
$\text{Ax}^2+\text{By}^2=1\ ...(1)$
Differentiating it with respect in x
$2\text{Ax}+2\text{By}\frac{\text{dy}}{\text{dx}}=0$
$\text{y}\frac{\text{dy}}{\text{dx}}=\frac{-2\text{Ax}}{2\text{B}}$
$\text{y}\frac{\text{dy}}{\text{dx}}=\frac{\text{Ax}}{\text{B}}$
Differentiating it with respect in x
$\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=-\frac{\text{A}}{\text{B}}$
$\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=\frac{\text{y}}{\text{x}}\frac{\text{dy}}{\text{dx}}$
Using equation (1)
$\text{x}\Big\{\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big\}=\text{y}\frac{\text{dy}}{\text{dx}}$
View full question & answer→Question 353 Marks
Solve the following differential equations:
$\sqrt{1+\text{x}^2}\ \text{dy}+\sqrt{1+\text{y}^2}\ \text{dx}=0$
Answer$\sqrt{1+\text{x}^2}\ \text{dy}+\sqrt{1+\text{y}^2}\ \text{dx}=0$
$\sqrt{1+\text{x}^2}\ \text{dy}=-\sqrt{1+\text{y}^2}\ \text{dx}$
$\int\frac{\text{dy}}{\sqrt{1+\text{y}^2}}=-\int\frac{\text{dx}}{\sqrt{1+\text{x}^2}}$
$\log|\text{y}+\sqrt{1+\text{y}^2}|=-\log|\text{x}+\sqrt{1+\text{x}^2}|=\log|\text{c}|$
$(\text{y}+\sqrt{1+\text{y}^2})(\text{x}+\sqrt{1+\text{x}^2})=\text{c}$
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Solve the following equation
$\text{xy dy}=(\text{y}-1)(\text{x}+1)\text{dx}$
Answer$\text{xy dy}=(\text{y}-1)(\text{x}+1)\text{dx}$
$\frac{\text{y}}{\text{y}-1}\text{dy}=\frac{\text{x}+1}{\text{x}}\ \text{dx}$
$\int\Big(1+\frac{1}{\text{y}-1}\Big)\text{dy}=\int\Big(1+\frac{1}{\text{x}}\Big)\text{dx}$
$\text{y}+\log|\text{y}-1|=\text{x}+\log|\text{x}|+\text{C}$
$\text{y}-\text{x}=\log|\text{x}|-\log|\text{y}-1|+\text{C}$
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Form the differential equation having $\text{y}=(\sin^{-1}\text{x})+\text{A}\cos^{-1}\text{x}+\text{B}$ where A and B are aribitrary constants, as its general solution.
Answer$\text{y}=(\sin^{-1}\text{x})+\text{A}\cos^{-1}\text{x}+\text{B}$
$\frac{\text{dy}}{\text{dx}}=2\sin^{-1}\text{x}\times\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)+\text{Ax}\Big(\frac{-1}{\sqrt{1-\text{x}^2}}\Big)=0$
$\sqrt{1-\text{x}^2}\frac{\text{dy}}{\text{dx}}=2\sin^{-1}\text{x}-\text{A}$
$\sqrt{1-\text{x}^2}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\frac{\text{dy}}{\text{dx}}\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)(-2\text{x})\Big(\frac{1}{\sqrt{1-\text{x}^2}}\Big)-0$
$(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{dx}^2}-2\text{x}\frac{\text{dy}}{\text{dx}}-2=0$
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Solve the following differential equations:
$\text{dy}+(\text{x}+1)(\text{y}+1)\text{dx}=0$
Answer$\text{dy}+(\text{x}+1)(\text{y}+1)\text{dx}=0$
$\text{dy}=-(\text{x}+1)(\text{y}+1)\text{dx}$
$\int\frac{\text{dy}}{\text{y}+1}=-\int(\text{x}+1)\text{dx}$
$\log|\text{y}+1|=-\frac{\text{x}^2}{2}-\text{x}+\text{C}$
$\log|\text{y}+1|+\frac{\text{x}^2}{2}+\text{x = C}$
View full question & answer→Question 393 Marks
Solve the following differential equation
$(\text{x}+2)\frac{\text{dy}}{\text{dx}}=\text{x}^2+3\text{x}+7$
Answer$(\text{x}+2)\frac{\text{dy}}{\text{dx}}=\text{x}^2+3\text{x}+7$
$\text{dy}=\Big(\frac{\text{x}^2+3\text{x}+7}{\text{x}+2}\Big)\text{dx}$
$\text{dy}-\Big(\text{x}+1+\frac{5}{\text{x}+2}\Big)\text{dx}$
$\int\text{dy}-\int\Big(\text{x}+1+\frac{5}{\text{x}+2}\Big)\text{dx}$
$\text{y}=\frac{\text{x}^2}{2}+\text{x}+5\log|\text{x}+2|+\text{C}$
$\text{x}\neq-2$
View full question & answer→Question 403 Marks
Represent the following families of curves by forming the corresponding differential equation:
$\text{y}^2=4\text{a}(\text{x}-\text{b})$
AnswerThe equation of the family of curves is
$\text{y}^2=4\text{a}(\text{x}-\text{b})\ ...(1)$
where a is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
$2\text{y}\frac{\text{dy}}{\text{dx}}=4\text{a}$
$\Rightarrow\text{y}\frac{\text{dy}}{\text{dx}}=2\text{a}$
Differentiating (2) with respect to x, we get
$\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=0$
It is the required differential equation.
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Solve the following differential equation
$\frac{\text{dy}}{\text{dx}}=\sin^2\text{y}$
AnswerWe have $\frac{\text{dy}}{\text{dx}}=\sin^2\text{y}$ $\Rightarrow\text{dx}=\frac{1}{\sin^2\text{y}}$ $\Rightarrow\text{dx}=\text{cosec}^2\text{y dy}$ Integrating both sides, we get $\int\text{dx}=\text{cosec}^2\text{y dy}$ $\Rightarrow\text{x}=-\cot\text{y}+\text{C}$$\Rightarrow\text{x}+\cot\text{y}=\text{C}$
Hence, $\Rightarrow\text{x}+\cot\text{y}=\text{C}$ is the required solution.
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Show that $\text{y}=\text{Ae}^{\text{bx}}$ is a solution of the differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1}{\text{y}}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2.$
AnswerWe have,
$\text{y}=\text{Ae}^{\text{bx}}\ ...(1)$
Differentiating both sides of equation (1) with respect to 3, we get
$\frac{\text{dy}}{\text{dx}}=\text{ABe}^{\text{Bx}}\ ...(2)$
Differentiating both sides of equation (2) with respect to 3, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{AB}^2\text{e}^{\text{Bx}}$
$=\frac{(\text{ABe}^{\text{Bx}^2})}{(\text{Ae}^{\text{Bx}})}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{1}{\text{y}}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$
Hence, the given function is the solution to the given differential equation.
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Solve the following equation:
$(\text{e}^\text{y}+1)\cos\text{x dx}+\text{e}^\text{y}\sin\text{x}\text{dy}=0$
Answer$(\text{e}^\text{y}+1)\cos\text{x dx}+\text{e}^\text{y}\sin\text{x}\text{dy}=0$
$(\text{e}^\text{y}+1)\cos\text{x dx}=-\text{e}^\text{y}\sin\text{x}\text{dy}$
$\int\frac{\cos\text{x}}{\sin\text{x}}\ \text{dx}=-\int\frac{\text{e}^\text{y}}{\text{e}^\text{y}+1}\ \text{dy}$
$\int\cot\text{x dx}=-\int\frac{\text{e}^\text{y}}{\text{e}^\text{y}+1}\ \text{dy}$
$\log|\sin\text{x}|=-\log|\text{e}^\text{y}+1|+\log\text|C|$
$\sin\text{x}=\frac{\text{C}}{\text{e}^\text{y}+1}$
$\sin\text{x}(\text{e}^\text{y}+1)=\text{C}$
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