Alternative English Year - 2017 Full Marks : 100
Class 12 Ncert Solution Maths Differential Equations
Previous Exercises
Exercise 9.1 &9.2
Next Exercises
Exercise 9.5
Exercise 9.6
Exercise 9.4
For each of the differential equations in Exercises 1 to 10, find the general solution:
1.\(\frac{dy}{dx}=\frac{1-\cos x}{1+ \cos x}\)
Sol: We have
\begin{aligned}
\frac{dy}{dx}&=\frac{1-\cos x}{1+ \cos x}\\
&= \frac{\sin^2{x\over2} + \cos^2{x\over2} - \cos^2{x\over2} + \sin^2{x\over2}}{\sin^2{x\over2} + \cos^2{x\over2} +\cos^2{x\over2} - \sin^2{x\over2}}\\
&= \frac{\sin^2{x\over2}+ \sin^2{x\over2}}{ \cos^2{x\over2} +\cos^2{x\over2}}\\
&= \frac{2\sin^2{x\over2}}{ 2\cos^2{x\over2}}\\
&= \tan^2{x\over2}\\
\frac{dy}{dx}&= (\sec^2{x\over2}-1)\\
\end{aligned}
Separating the variables of the above eqaution we get
$$dy=(\sec^2{x\over2}-1)dx$$
Now, integrating both sides of the above equation, we get
\begin{aligned}
\int dy&= \int(\sec^2{x\over2}-1)dx\\
y&= \frac{\tan^2{x\over2}}{1\over2} - x + C\\
y &= 2\tan^2{x\over2} -x + C
\end{aligned}
which is the general solution of the given differential equation
2. \(\frac{dy}{dx}=\sqrt{4-y^2}\)
Sol: We have
$$\frac{dy}{dx}=\sqrt{4-y^2}$$
Separating the variables of the above equation we get
$$\frac{dy}{\sqrt{4-y^2}} = dx$$
Integrating both sides of the equation, we get
\begin{aligned}
\int \frac{dy}{\sqrt{4-y^2}} &= \int dx\\
\int \frac{dy}{\sqrt{2^2-y^2}}& = \int dx\\
\sin^{-1}\frac{y}{a}&= x+ C\\
\frac{y}{a} &= sin(x+C)\\
y&=a\sin(x+C)\\
\end{aligned}
which is the general solution of the given differential equation.
3. \(\frac{dy}{dx} + y = 1(y\neq1)\)
Sol: We have
\begin{aligned}
\frac{dy}{dx} + y &= 1(y\neq1) \\
\frac{dy}{dx} &=1-y\\
-\frac{dy}{y-1}&=x\\
\end{aligned}
Separating the variables in the above equation, we get
$$-\frac{dy}{y-1}=x $$
Integrating both sides of the equation we get
\begin{aligned}
\int \frac{dy}{y-1}&=-\int x\\
log|y-1| &=-x + C\\
y-1 &=e^{-x+ C}\\
y&=1+e^{-x}.e^C\\
y&=1+Ae^{-x}\\
\end{aligned}
4. \(\sec^2 x\tan y\ dx + \sec^2 y \tan x\ dy = 0\)
Sol: We have
$$ \sec^2 x\tan y\ dx + \sec^2 y \tan x\ dy = 0$$
Separating the variables of the above equation we get
$$ \frac{\sec^2x}{\tan x}\ dx + \frac{\sec^2y}{\tan y}\ dy = 0$$
Integrating both sides of the equation we get
\begin{aligned}
\int \frac{\sec^2x}{\tan x}\ dx + \int \frac{\sec^2y}{\tan y}\ dy = \int0\\
\int \frac{d(\tan x)}{\tan x}\ dx + \int \frac{d(\tan y)}{\tan y}\ dy = \int0\\
log|\tan x| + log|\tan y| = \log C\\
log|\tan x\tan y| = \log C\\
\tan x\tan y = C\\
\end{aligned}
which is the general solution of the given differential equation
5.\((e^x + e^{-x})dy - (e^x - e^{-x})dx = 0\)
Sol: We have
$$ (e^x + e^{-x})dy - (e^x - e^{-x})dx = 0 $$
Separating the variables of the above equation we get
$$ dy = \frac{e^x + e^{-x}}{e^x - e^{-x}}dx$$
Integrating both sides of the above equation, we get
\begin{aligned}
\int dy = \int\frac{e^x + e^{-x}}{e^x - e^{-x}}dx\\
\int dy = \int\frac{d(e^x - e^{-x})}{e^x - e^{-x}}dx\\
y = \log |e^x - e^{-x}| + C
\end{aligned}
which is the general solution of the given differential equation.
6. \(\frac{dy}{dx}=(1+ x^2)(1+ y^2)\)
Sol: We have $$ \frac{dy}{dx}=(1+ x^2)(1+ y^2)$$
Separating the variables of the above equation we get
$$\frac{dy}{1+ y^2} = (1+ x^2)dx$$
Integrating both sides of the above equation we get
\begin{aligned}
\int \frac{dy}{1+ y^2} = \int (1+ x^2)dx\\
\int \frac{dy}{1+ y^2} = \int dx+ \int x^2dx\\
\tan^{-1}y = x+ \frac{x^3}{3} + C
\end{aligned}
which is the general solution of the given differential equation
7. \(y \log y\ dx - x\ dy =0\)
Sol: We have $$ y \log y\ dx - x\ dy =0$$
Separating the variables of the above equation we get
$$ \frac{dy}{y\log y} =\frac{dx}{x}$$
Integrating both sides of the above equation,we get
\begin{aligned}
\int \frac{dy}{y\log y} &=\int \frac{dx}{x}\\
\int \frac{d(\log y)}{\log y} &=\int \frac{dx}{x}\\
\log |\log y| &= \log x + \log C\\
\log |\log y| &= \log xC\\
\log y &=xC\\
y &= e^{xC}\\
\end{aligned}
which is the genral solution of the given differential equation
8. \(x^5\frac{dy}{dx} = - y^5\)
Sol: We have $$ x^5\frac{dy}{dx} = - y^5 $$
Separating the variables of the above equation we get
$$\frac{dy}{y^5} + \frac{dx}{x^5}=0$$
Integrating both sides of the above equation we get
\begin{align}
\int \frac{dy}{y^5} + \int \frac{dx}{x^5} = A\\
\\
\frac{y^{-5+1}}{-5+1} + \frac{x^{-5+1}}{-5+1}= A\\
\\
\frac{y^{-4}}{-4} + \frac{y^-4}{-4} =A\\
y^{-4} + x^{-4} = -4A\\
y^{-4} + x^{-4} = C\\
\end{align}
which is the required general solution of the given differential equation
9. \(\frac{dy}{dx} = \sin^{-1} x\)
Sol: We have $$ \frac{dy}{dx} = \sin^{-1} x$$
Separating the variables in the above equation we have
$$ dy = \sin^{-1} x\ dx$$
Integrating both sides of the above equation we get
\begin{aligned}
\int dy &= \int \sin^{-1} x\ dx\\
y &= \sin^{-1}x\int 1dx -\int \frac{1}{\sqrt{1-x^2}}\ dx\ (\text{Integrating by parts})\\
y&= x\sin^{-1} x + \int\frac{-2x}{2\sqrt{1-x^2}}\ dx\\
y&= x\sin^{-1} x + \int d(\sqrt{1-x^2})\\
y&= x\sin^{-1} x + \sqrt{1-x^2} + C
\end{aligned}
which is the general solution of the given differential equation.
10 \(e^x \tan y \ dx + (1- e^x) \sec^2y\ dy =0\)
Sol: We have $$e^x \tan y \ dx + (1- e^x) \sec^2y\ dy =0$$
Separating the variables, the above equation can be written as
$$\left(\frac{e^x}{1-e^x}\right)dx + \left(\frac{\sec^2y}{\tan y}\right)dy=0 $$
Integrating both sides of the above equation we get
\begin{aligned}
\int \left(\frac{e^x}{1-e^x}\right)dx +\int \left(\frac{\sec^2y}{\tan y}\right)dy&=\int 0\\
\int- \left(\frac{-e^x}{1-e^x}\right)dx +\int \left(\frac{\sec^2y}{\tan y}\right)dy&=\int 0\\
\int \left(\frac{d(1-e^x)}{1-e^x}\right)dx +\int \left(\frac{d(\tan y)}{\tan y}\right)dy&=\int 0\\
-log|1-e^x| + log|\tan y| &= \log C\\
\log\left(\frac{\tan y}{1-e^x}\right) &= \log C\\
\tan y &= C(1-e^x)\\
\end{aligned}
which is the general solution of the given differential equation.
For each of the following of the differential equations in Exercises 11 to 14, find a particular solution satisfying the given condition:
11. \(x^3 + x^2 + x + 1)\frac{dy}{dx} = 2x^2 + x: y =1\) when x =0
Sol: We have $$ (x^3 + x^2 + x+ 1)\frac{dy}{dx} = 2x^2 + x$$
Separating the variables the above equation can be written as
$$ dy = \frac{2x^2 + x}{x^3 + x^2 + x+ 1}\ dx$$
Integrating both sides of the equation we get
\begin{align}
\int dy & = \int \frac{2x^2 + x}{x^3 + x^2 + x+ 1}\ dx\\
y&=\int \frac{2x^2 + x}{(x^2 +1)( x+ 1)}\ dx\tag 1\\
\\
Let\ \frac{2x^2 + x}{(x^2 +1)( x+ 1)} &=\frac{A}{x+1} + \frac{Bx+C}{x^2+1}\\
\frac{2x^2 + x}{(x^2 +1)( x+ 1)}&= \frac{A(x^2+1) +(Bx+C)(x+1)}{(x+1)(x^2+1)}\\
2x^2 + x&=A(x^2+1) +(Bx+C)(x+1)\\
2x^2 + x&=Ax^2 +Bx^2+ Bx + Cx + A+C\\
\\
\text{Equating the coefficients of corresponding terms we get}\\
A+B = 2\\
B+C = 1\\
A+C = 0\\
\text{Solving the above three equations we get}\\
A=\frac{1}{2},\ \ B=\frac{3}{2},\ \ C=\frac{-1}{2}\\
\text{Substituting these values in eq(1) we get}\\
y=\int \frac{1}{2(x+1)}\ dx + \int \frac{3x-1}{2(x^2+1)}\ dx\\
y=\int \frac{1}{2(x+1)}\ dx + \int \frac{3x}{2(x^2+1)}\ dx -\int \frac{1}{2(x^2+1)}\ dx\\
y=\int \frac{1}{2(x+1)}\ dx + \frac{3}{4}\int \frac{2x}{(x^2+1)}\ dx -\frac{1}{2}\int\frac{1}{(x^2+1)}\ dx\\
y= \frac{1}{2}log|x+1| + \frac{3}{4}log|x^2+1| -\frac{1}{2}\tan^{-1}x+C \tag 1\\
\end{align}
Given that y=1 when x=0. Substituting these values in above equation we get
\begin{align}
1&= \frac{1}{2}log|0+1| + \frac{3}{4}log|0^2+1| -\frac{1}{2}\tan^{-1}0+C\\
C&=1
\end{align}
Substituting the value of C in eq (1) we get
\begin{align}
y&=\frac{1}{2}log|x+1| + \frac{3}{4}log|x^2+1| -\frac{1}{2}\tan^{-1}x+1\\
y&=\frac{1}{4}log|x+1|^2+ \frac{1}{4}log|x^2+1|^3 -\frac{1}{2}\tan^{-1}x+1\\
y&=\frac{1}{4}\log\left[(x+1)^2(x^2+1)^3\right]-\frac{1}{2}\tan^{-1}x+1\\
\end{align}
12. \(x(x^2 - 1)\frac{dy}{dx} =1: y=0\) when x =2
Sol: We have
$$ x(x^2-1)\frac{dy}{dx} =1 $$
Separating the variables the above equation can be written as
$$dy = \frac{dx}{x(x+1)(x-1)}$$
Integrating both sides of the above equation, we get
\begin{align}
\int dy &= \int \frac{dx}{x(x+1)(x-1)}\\
\int dy &=\int \left(\frac{-1}{x} + \frac{1}{2(x+1)} + \frac{1}{2(x-1)}\ dx\right)\\
y&= -\log x +\frac{1}{2}\log+1) + \frac{1}{2}\log-1) + \log k\\
y&=\frac{1}{2}\log\left[\frac{k^2(x+1)(x-1)}{x^2}\right]\tag 1\\
\end{align}
Given that y=0 when x=2. Substituting these values in eq (1) we get
\begin{align}
0&=\frac{1}{2}\log\left[\frac{k^2(2+1)(2-1)}{2^2}\right]\\\
\frac{3k^2}{4}&= e^0\\
k^2 &=\frac{4}{3}\\
\end{align}
Substituting this value of k in eq (1) we get
\begin{align}
y&=\frac{1}{2}\log\left[\frac{4(x+1)(x-1)}{3x^2}\right]
\end{align}
Class 12 Ncert Solution Maths Differential Equations
13. \(\cos\left(\frac{dy}{dx}\right)=a(a\in R)\);y =1 when x =0
Sol: We have
$$\cos\left(\frac{dy}{dx}\right)=a$$
Separating variables the above equation can be written as
\begin{align}
\frac{dy}{dx}&=\cos^{-1}a\\
dy&=\cos^{-1}a\ dx \tag 1\\
\end{align}
Integrating both sides of eq (1) we get
\begin{align}
\int dy&=\int \cos^{-1}a\ dx
y &= x\cos^{-1}a + C\tag 2\\
\end{align}
Given that y=1 when x = 0. Putting these values in eq (2) we get
\begin{align}
1 &= 0\times\cos^{-1}a + C\\
C&=1\\
\end{align}
Substituting the value of C in eq (2) we get
\begin{align}
y &= x\cos^{-1}a + 1\\
y-1&=x\cos^{-1}a\\
\frac{y-1}{x} &= \cos^{-1}a\\
\cos \frac{y-1}{x} &= a\\
\end{align}
14. \(\frac{dy}{dx} =y\tan x\) ;y=1 when x =0
Sol: We have $$ \frac{dy}{dx} =y\tan x$$
Separating the variables the baove equation can be written as
$$ \frac{dy}{y}=\tan x\ dx $$
Integrating both sides of the above equation we get
\begin{align}
\int \frac{dy}{y} &= \int \tan x \ dx\\
\log y&=\log (\sec x) + log C\\
\log y &=\log (C\sec x)\\
y&= C\sec x\tag 1\\
\end{align}
Given that y =1 when x =0. Substituting these values in eq(1) we get
\begin{align}
1&=C\sec 0\\
C&=1\\
\end{align}
Substituting this value of C in (1) we get
$$y=\sec x$$
15. Find the equation of a curve passing through the point (0,0) and whose differential equation is \(y^{'}=e^x\sin x\).
Sol: The given differential equation is
$$ y^{'}=e^x\sin x $$
$$\frac{dy}{dx} = e^x\sin x$$
$$dy= e^x\sin x dx$$
Integrating both sides of the equation, we get
\begin{align}
\int dy &= \int e^x\sin x dx\\
y&= \sin x\ e^x - \int e^x \cos x\ dx \ (\text{Integration by parts})\\
y&= \sin x \ e^x - \left[\cos x e^x -\int(-\sin x) e^x\ dx\right]\\
y&= \sin x \ e^x - \left[\cos x e^x +\int(\sin x) e^x\ dx\right]\\
y&= \sin x \ e^x - \cos x e^x -\int(\sin x) e^x\ dx\\
y&= e^x(\sin x - \cos x) - y\\
y+y&= e^x(\sin x - \cos x) \\
2y&= e^x(\sin x - \cos x) \\
y&= \frac{1}{2}e^x(\sin x - \cos x) + C\tag 1\\
\end{align}
Substituting x=0 and y=0 in the above equation we get
\begin{aligned}
0 = \frac{1}{2}e^0(\sin 0 - \cos 0) + C\\
0 = \frac{1}{2}1(- 1) + C\\
C = -\frac{1}{2}\\
\end{aligned}
Substituting the value of C in equation(1) we get
\begin{align}
y &= \frac{1}{2}e^x(\sin x - \cos x) + \frac{1}{2}\\
2y&= \frac{1}{2}e^x(\sin x - \cos x) + 1\\
2y-1&= \frac{1}{2}e^x(\sin x - \cos x) \\
\end{align}
which is the required equation of a curve passing throught the point (0,0).
16. For the differential equation \(xy\frac{dy}{dx}= (x+2)(y+2)\), find the solution curve passing through the point (1,-1).
Sol: We have $$ xy\frac{dy}{dx}= (x+2)(y+2) $$
Separating the variables the above equation can be written as
$$ \frac{ydy}{y+2}=\frac{x+2}{x}\ dx$$
Integrating both sides of the above equation, we get
\begin{align}
\int \frac{ydy}{y+2}=\int \frac{x+2}{x}\ dx\\
\int \frac{y+2-2}{y+2}\ dy &=\int \frac{x+2}{x}dx\\
\int1.dy -2 \frac{1}{y+2}\ dy &=\int dx+ 2 \frac{1}{x}\ dx\\
y - 2\log|y+2| &= x + 2 \log x + C\tag 1\\
\end{align}
The curve passes through the point (1, -1). Substituting x =1 and y = -1 in eq (1) we get
\begin{align}
-1 - 2\log|-1+2| &= 1 + 2\log 1 + C\\
C&= -2\ \text{since log 1 =0}\\
\end{align}
Substituting the value of C in eq (1) we get
\begin{align}
y - 2\log|y+2| &= x + 2 \log x + -2\\
y-x+2 &= log(y+2)^2 + log x^2\\
y+x-2 &= log (x^2(y+2)^2)\\
\end{align}
which is the required equation of the curve passing through the point (1, -1)
17. Find the equation of a curve passing through the point (0,-2) given that at any point (x,y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.
Sol: Slope of a tangent is = \(\frac{dy}{dx}\)
Given that \(y\frac{dy}{dx}=x\)
Separating the variables the above equation can be written as
$$ y\ dy= x\ dx$$
Integrating both sides of the equation we get
\begin{align}
\int y\ dy &= \int x dx\\
\frac{y^2}{2} &= \frac{x^2}{2} + C \tag 1\\
\end{align}
Given that the curve passes throught the point(0, -2)
Substituting x = 0 and y =-2 in eq (1) we get
\begin{align}
\frac{(-2)^2}{2} &= \frac{0^2}{2} + C\\
\frac{4}{2} &= C\\
C &= 2\\
\end{align}
Substituting the value of C in eq (1) we get
\begin{align}
\frac{y^2}{2} &= \frac{x^2}{2} + C\\
\frac{y^2}{2} &= \frac{x^2}{2} + 2\\
\frac{y^2}{2}& = \frac{x^2 +4}{2}\\
y^2 = x^2 + 4\\
y^2 - x^2 =4\\
\end{align}
which is the required equation of the curve passing through the point (0,-2)
18. At any point (x,y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (-4,-3). Find the equation of the curve given that it passes through (-2,1).
Sol: The slope of the line passing through the point (x,y) and (-4,-3) is = \(\frac{y+3}{x+4}\)
We know slope of the tangent =\(\frac{dy}{dx}\)
Given that
\begin{align}
\frac{dy}{dx} = 2\frac{y+3}{x+4}\\
\frac{dy}{y+4} = 2(x+3)dx\\
\end{align}
Integrating both sides of the equation
\begin{align}
\int \frac{dy}{y+3} &= \int frac{2}{x+4}dx \\
\log |y+3| &= 2\log|x+4| + \log C\\
\log |y+3| &= \log C(x+4)^2\\
y+3&=C(x+4)^2\\tag 1\
\end{align}
The Curve passes through the point (-2,1), Substituting x=-1 and y=1 in eq(1) we get
\begin{align}
1+3&=C(-2+4)^2
4&= C\times 4
C&=1
\end{align}
Substituting C=1 in eq (1) we get
$$ y+3 = (x+4)^2 $$
Which is the general solution of the given differential equation.
19. The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.
Sol: Let r be the radius of the spherical baloon and V be the volume.
Therefore, $$\frac{dV}{dt} = C$$
$$dV =Cdt$$
Integrating both sides we get
\begin{align}
\int dV &= \int Cdt\\
V&= Ct + A, \text{where A is a constant}\\
\frac{4\pi r^3}{3}&=Ct + A \tag 1\\
\end{align}
Given that at t =0 , r =3
putting these values in eq (1) we get
\begin{align}
\frac{4\pi 3^3}{3}&=C\times0 + A\\
A&=36\pi \\
\end{align}
equation (1) now becomes
\begin{align}
\frac{4\pi r^3}{3}&=Ct + 36\pi\tag 2
\end{align}
When t = 3 sec , r= 6. Putting these values in eq (2) we get
\begin{align}
\frac{4\pi 6^3}{3}&=C\times 3+ 36\pi\\
288\pi&= 3C + 36\pi\\
C&= 84\pi\\
\end{align}
Thus, eq(2) becomes
\begin{align}
\frac{4\pi r^3}{3}&=84\pi t + 36\pi\\
4\pi r^3 &= 3\times 4\pi(21t + 9)\\
r^3&=62t + 27\\
r&= (62t+ 27)^\frac{1}{3}\\
\end{align}
This is the required radius of the spherical baloon after t seconds
20.In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 double itself in 10 years(log 2 = 0.6391)
Sol: Let p be the principal so given that
\begin{align}
\frac{dp}{dt} &= r%\times p \\
\frac{dp}{dt} &= \frac{r}{100}p\\
\frac{dp}{p} &= \frac{r}{100}dt\\
\end{align}
Integrating both sides of the equation we get
\begin{align}
\int \frac{dp}{p} = \int \frac{r}{100} \ dt \\
log|p|&=\frac{rt}{100} + \log C\\
\log p - \log C= \frac{rt}{100} \\
log|\frac{p}{C}| =\frac{rt}{100}\\
p=Ce^{\frac{rt}{100}}\tag 1\\
\end{align}
When t=0, p = 100. Substituting these values in eq (1) we get
\begin{align}
100 &= Ce^{0} \\
C&=100\\
\end{align}
putting this value of C in eq (1) we get
\begin{align}
p=100e^{\frac{rt}{100}}\tag 2\\
\end{align}
When t=10, p= 200. Substituting these values in eq (2) we get
\begin{align}
200&=100e^{\frac{r\times10}{100}}\\
2&= e^{\frac{r}{10}}\\
log 2&=\frac{r}{10}\\
r&=10\log 2\\
r&=10 \times0.6931\\
r&= 6.931\\
\end{align}
Thus the value of r is 6.93%
Class 12 Ncert Solution Maths Differential Equations
Sol: Let p be the principal. Given that
\begin{align}
\frac{dp}{dt} &= \frac{5}{100}p\\
\frac{dp}{dt} &= \frac{p}{20}\\
\frac{dp}{p} &=\frac{dt}{20}\\
\end{align}
Integrating both sides of the equation we get
\begin{align}
log|p| &= \frac{t}{20} + \log C \\
\log|p| -\log C &= \frac{t}{20}\\
\log \frac{p}{C} &= \frac{t}{20}\\
p&=Ce^{\frac{t}{20}}\tag 1\\
\end{align}
when t = 0, p= 1000. Substituting these value in the above equation we get
\begin{align}
1000&=Ce^{\frac{0}{20}}\\
C=1000\\
\end{align}
Putting this value of C in eq(1) we get
\begin{align}
p&=1000e^{\frac{t}{20}}\tag2\\
\end{align}
When t =10 eq(2) becomes
\begin{align}
p&= 1000e^{\frac{5}{20}}\\
p&= 1000e^{\frac{1}{4}}\\
p&=1000 e^{0.5}\\
p&= 1000 \times 1.648\\
p&= 1648\\
\end{align}
Thus, after 10 years the deposited amount will become Rs 1648.
22. In a culture, the bacteria count is 1,00,000. The number is increased by 10 % in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?
Sol: Let the number of bacteria at any time t be n.
Thus given that
\begin{align}
\frac{dn}{dt} \alpha\ n\\
\frac{dn}{dt} &= kn\\
\frac{dn}{n}&=kdt\\
\end{align}
Integrating the aboce equation we get
\begin{align}
log|n| &=kt + C \tag 1\\
\end{align}
When t=0, n=100000. Substituting these values in eq (1) we get
\begin{align}
\log 100000 &= k\times 0 + C\\
C &= \log100000\\
\end{align}
Putting this value of C in eq (1) we get
\begin{align}
log|n| &=kt + \log 100000\tag 2\\
\end{align}
The number increases by 10 % in 2 hours
So t = 2 and n = 100000 + 10% of 100000
therefore, n = 110000
Substituting these values in eq(2) we get
\begin{align}
\log110000 &=2k + \log100000\\
\log110000-\log 100000 &= 2k\\
\log \left(\frac{110000}{100000}\right)& = 2k\\
k&= \frac{1}{2}\log\left(\frac{11}{10}\right)\\
\end{align}
Substituting this value of k in eq (2), we get
\begin{align}
log|n| &=\frac{1}{2}\log\left(\frac{11}{10}\right)t + \log 100000\\
\end{align}
Now when n =200000 eq (3) becomes
\begin{align}
\log 200000 &= \frac{1}{2}\log\left(\frac{11}{10}\right)t - \log 100000\\
\log 200000 - \log 100000 &=\frac{1}{2}\log\left(\frac{11}{10}\right)t\\
\log\left(\frac{200000}{100000}\right)&= \frac{1}{2}\log\left(\frac{11}{10}\right)t\\
\log 2 &= \frac{1}{2}\log\left(\frac{11}{10}\right)t\\
t &=\frac{log2}{ \frac{1}{2}\log\left(\frac{11}{10}\right)}\\
t&=\frac{2log2}{ \log\left(\frac{11}{10}\right)}\\
\end{align}
23. The general solution of the differential equation\(\frac{dy}{dx}=e^{x+y}\) is
(A) \(e^x + e^{-y} = C\) (B) \(e^x + e^{y} = C\)
(C) \(e^{-x} + e^{y} = C\) (D) \(e^{-x} + e^{-y} = C\)
Sol: The given differential equation can be written as
$$ \frac{dy}{e^y} =e^x\ dx$$
Integrating both sides we get
\begin{align}
\int \frac{dy}{e^y} &=\int e^x\ dx\\
\int e^{-y} &=\int e^x\ dx\\
-e^{-y} + C&=e^x \\
e^x +e^{-y}&=C\\
\end{align}
So the correct answer is option (A).
Class 12 Ncert Solution Maths Differential Equations
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Exercise 9.1 &9.2
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Exercise 9.5
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