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Maths (commerce) Solve the Following Question.(5 Marks)

Differential Equation and Applications (p-1) · Maharashtra Board STD 12 Commerce / Arts (MSBSHSE - English Medium). 21 questions.

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Solve the Following Question.(5 Marks)

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21 questions · self-marked practice — reveal the answer and mark yourself.

Question 15 Marks
The rate of growth of the population is proportional to the number present. If the population doubled in the last 25 years and the present population is 1 lakh, when will the city have a population of 400000 ?
Answer
Let $P$ be the population at time $t$ years.
Then the rate of growth of the population is $\frac{d P}{d t}$ which is proportional to $P$.
$
\begin{aligned}
& \therefore \frac{d P}{d t} \propto P \\
& \therefore \frac{d P}{d t}= kP , \text { where } k \text { is a constant } \\
& \therefore \frac{d P}{P}= kdt
\end{aligned}
$
On integrating, we get
$
\begin{aligned}
& \int \frac{d P}{P}=k \int d t \\
& \therefore \log P = kt + c
\end{aligned}
$
The population doubled in 25 years and present population is $1,00,000$.
$\therefore$ initial population was 50,000
i.e. when $t=0, P=50000$
$
\therefore \log 50000= k \times 0+ c
$
$
\therefore c =\log 50000
$
$
\therefore \log P=k t+\log 50000
$
When $t =25, P =100000$
$
\therefore \log 100000= k \times 25+\log 50000
$
$
\therefore 25 k =\log 100000-\log 50000=\log \left(\frac{100000}{50000}\right)
$
$\therefore k =\frac{1}{25} \log 2$
$
\begin{aligned}
& \therefore \log P=\frac{t}{25} \log 2+\log 50000 \\
& \text { If } P=400000, \text { then } \\
& \log 400000=\frac{t}{25} \log 2+\log 50000 \\
& \therefore \log 400000-\log 50000=\frac{t}{25} \log 2 \\
& \therefore \log \left(\frac{400000}{50000}\right)=\log (2)^{t / 25} \\
& \therefore \log 8=\log (2)^{t / 25} \\
& \therefore 8=(2)^{t / 25} \\
& \therefore(2)^{t / 25}=(2)^3 \\
& \therefore \frac{t}{25}=3 \\
& \therefore t =75
\end{aligned}
$
$\therefore$ the population will be 400000 in $(75-25)=50$ years.
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Question 25 Marks
Solve y $\log y \frac{d x}{d y}+ x -\log y =0$
Answer
$
\begin{aligned}
& y \log y \cdot \frac{d x}{d y}+x-\log y=0 \\
& \therefore y \log y \cdot \frac{d x}{d y}=\log y-x \\
& \therefore \frac{d x}{d y}=\frac{\log y-x}{y \log y} \\
& \therefore \frac{d x}{d y}=\frac{1}{y}-\frac{x}{y \log y} \\
& \therefore \frac{d x}{d y}+\frac{x}{y \log y}=\frac{1}{y}
\end{aligned}
$
This is the linear differential equation of the form $\frac{d x}{d y}+P x=Q$, where $P=\frac{1}{y \log y}$ and $Q=\frac{1}{y}$
$
\begin{aligned}
\therefore \text { I.F. } & =e^{\int P d y}=e^{\int \frac{1}{y \log y} d y} \\
& =e^{\int \frac{(1 / y)}{\log y} d y}=e^{\log |\log y|}=\log y
\end{aligned}
$
$\therefore$ the solution of (1) is given by
$
\begin{aligned}
& x \cdot(\text { I.F. })=\int Q \cdot(\text { I.F. }) d y+c_1 \\
\therefore x \cdot \log y & =\int \frac{1}{y} \cdot \log y d y+c_1 \\
\therefore \quad & (\log y) \cdot x=\int \frac{\log y}{y} d y+c_1
\end{aligned}
$
Put $\log y=t \quad \therefore \frac{1}{y} d y=d t$
$
\begin{aligned}
& \therefore(\log y) \cdot x=\int t d t+c_1 \\
& \therefore x \log y=\frac{t^2}{2}+c_1
\end{aligned}
$
$
\begin{aligned}
& \therefore x \log y=\frac{1}{2}(\log y)^2+c_1 \\
& \therefore 2 x \log y=(\log y)^2+c, \text { where } c=2 c_1
\end{aligned}
$
This is the general solution.
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Question 35 Marks
Solve $y d x-x d y=-\log x d x$
Answer
$
\begin{aligned}
& y d x-x d y=-\log x d x \\
& \therefore ydx - xdy +\log xdx =0 \\
& \therefore xdy =( y +\log x ) dx \\
& \therefore \frac{d y}{d x}=\frac{y+\log x}{x}=\frac{y}{x}+\frac{\log x}{x} \\
& \therefore \frac{d y}{d x}-\frac{1}{x} \cdot y=\frac{\log x}{x} \\
\end{aligned}
$
This is the linear differential equation of the form
$
\begin{aligned}
& \frac{d y}{d x}+P \cdot y=Q, \text { where } P=-\frac{1}{x} \text { and } Q=\frac{\log x}{x} \\
& \begin{aligned}
\therefore \text { I.F. } & =e^{\int P d x}=e^{\int-\frac{1}{x} d x}=e^{-\log x} \\
& =e^{\log (x)^{-1}}=\frac{1}{x}
\end{aligned}
\end{aligned}
$
$\therefore$ the solution of (1) is given by
$
\begin{aligned}
& y \cdot\left(\text { I.F.) }=\int Q \cdot(\text { I.F.) } d x+c\right. \\
\therefore & y \cdot \frac{1}{x}=\int \frac{\log x}{x} \cdot \frac{1}{x} d x+c \\
\therefore & \frac{y}{x}=\int \frac{\log x}{x^2} d x+c \\
\therefore & \frac{y}{x}=(\log x) \int x^{-2} d x-\int\left[\frac{d}{d x}(\log x) \int x^{-2} d x\right] d x+c \\
\therefore & \frac{y}{x}=(\log x) \cdot \frac{x^{-1}}{-1}-\int \frac{1}{x} \cdot \frac{x^{-1}}{-1} d x+c \\
\therefore & \frac{y}{x}=-\frac{\log x}{x}+\int x^{-2} d x+c \\
\therefore & \frac{y}{x}=-\frac{\log x}{x}+\frac{x^{-1}}{-1}+c \\
\therefore & \frac{y}{x}=-\frac{\log x}{x}-\frac{1}{x}+c
\end{aligned}
$
$
\begin{aligned}
& \therefore y=-\log x-1+c x \\
& \therefore y=c x-(1+\log x)
\end{aligned}
$This is the general solution.
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Question 45 Marks
$y \log y \frac{d x}{d y}=\log y - x$
Answer
$y \log y \frac{d x}{d y}=\log y-x$
$y \log y \frac{d x}{d y}+x=\log y$
$\therefore \frac{d x}{d y}+\frac{1}{y \log y} x=\frac{1}{y}$
The given equation is of the form $\frac{d x}{d y}+p x=Q$
where, $P=\frac{1}{y \log y}$ and $Q=\frac{1}{y}$
$\therefore I . F .=e^{\int^{p d y}=e^{\int^{\frac{1}{y \log y}} d y}=e^{\log \mid \log y}=\log y}$
$\therefore$ Solution of the given equation is
$x(I . F)=.\int Q(I . F) d y+.c_1$
$\therefore x \cdot \log y=1 y \int \log y d y+c_1$
In R. H. S., put $\log y=t$
Differentiating w.r.t. $x$, we get
$
\frac{1}{y} d y=d t
$
$\begin{aligned} & \therefore x \log y=t d t \int+c_1=\frac{t^2}{2}+c_1 \\ & \therefore x \log y=\frac{(\log y)^2}{2}+c_1 \\ & \therefore 2 x \log y=(\log y)^2+c \ldots\left[2 c_1=c\right] \\ & x \log y=\frac{1}{2}(\log y)^2+c\end{aligned}$
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Question 55 Marks
Solve $y d x-x d y+\log x d x=0$
Answer
$
\begin{aligned}
& y d x-x d y+\log x d x=0 \\
& \therefore(y+\log x) d x=x d y \\
& \therefore \frac{y+\log x}{x}=\frac{d y}{d x} \\
& \therefore \frac{y}{x}+\frac{\log x}{x}=\frac{d y}{d x} \\
& \therefore \frac{d y}{d x}-\frac{1}{x} \cdot y=\frac{\log x}{x}
\end{aligned}
$
This is a linear differential equation of the form
$
\frac{d y}{d x}+P y=Q \text {, where } P=-\frac{1}{x}, Q=\frac{\log x}{x}
$
$
\begin{aligned}
\therefore \text { I.F. } & =e^{\int P d x}=e^{\int-\frac{1}{x} d x}=e^{-\int \frac{1}{x} d x} \\
& =e^{-\log x}=e^{\log \left(\frac{1}{x}\right)}=\frac{1}{x}
\end{aligned}
$
$\therefore$ the solution of (1) is given by
$
y \cdot\left(\text { I.F.) }=\int Q \cdot \text { (I.F.) } d x+c\right.
$
$
\begin{aligned}
\therefore \frac{y}{x} & =\int \frac{\log x}{x} \cdot \frac{1}{x} d x+c \\
& =\int \log x \cdot x^{-2} d x+c
\end{aligned}
$
$
\begin{aligned}
& =(\log x) \int x^{-2} d x-\int\left[\left\{\frac{d}{d x}(\log x)\right\} \int x^{-2} d x\right] d x \\
& =(\log x)\left(\frac{x^{-1}}{-1}\right)-\int \frac{1}{x} \cdot\left(\frac{x^{-1}}{-1}\right) d x+c
\end{aligned}
$
$
=-\frac{\log x}{x}+\int x^{-2} d x+c
$
$
\begin{aligned}
& \therefore \frac{y}{x}=-\frac{\log x}{x}-\frac{1}{x}+c \\
& \therefore y=c x-(1+\log x)
\end{aligned}
$
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Question 65 Marks
Solve $x^2 y d x-\left(x^3+y^3\right) d y=0$
Answer
$
\begin{aligned}
& x^2 y d x-\left(x^3+y^3\right) d y=0 \\
& \therefore x^2 y d x-\left(x^3+y^3\right)=d y \\
& \therefore \frac{d y}{d x}=\frac{x^2 y}{x^3+y^3} \ldots \text { (i) }
\end{aligned}
$
Put $y=t x$
Differentiating w.r.t. $x$, we get
$
\frac{d y}{d x}=t+x \frac{d t}{d x}
$
Substituting (ii) and (iii) in (i), we get
$
\begin{aligned}
& t+x \frac{d t}{d x}=\frac{x^2 \cdot t x}{x^3+t^3 x^3} \\
& \therefore t+x \frac{d t}{d x}=\frac{x^3 \cdot t}{x^3\left(1+t^3\right)} \\
& \therefore x \frac{d t}{d x}=\frac{t}{1+t^3}-t \\
& \therefore x \frac{d t}{d x}=\frac{t-t-t^4}{1+t^3} \\
& \therefore x \frac{d t}{d x}=\frac{-t^4}{1+t^3} \\
& \therefore \frac{1+t^3}{t^4} d t=-\frac{d x}{x}
\end{aligned}
$
Integrating on both sides, we get
$
\begin{aligned}
& \int \frac{1+t^3}{t^4} d t=-\int \frac{1}{x} d x \\
& \therefore \int\left(\frac{1}{t^4}+\frac{1}{t}\right) d t=-\int \frac{1}{x} d x \\
& \therefore \int t^{-4} d t+\int \frac{1}{t} d t=-\int \frac{1}{x} d x \\
& \therefore \frac{t^3}{-3}+\log |t|=-\log |x|+c \\
& \therefore-\frac{1}{3 t^3}+\log |t|=-\log |x|+c \\
& \therefore-\frac{1}{3} \cdot \frac{1}{\left(\frac{y}{x}\right)^3}+\log \left|\frac{y}{x}\right|=-\log |x|+c \\
& \therefore \frac{x^3}{3 y^3}+\log |y|-\log |x|=-\log |x|+c \\
& \therefore \log |y|-\frac{x^3}{3 y^3}=c
\end{aligned}
$
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Question 75 Marks
Solve $y^2 d x+\left(x y+x^2\right) d y=0$
Answer
$
\begin{aligned}
& y 2 d x+\left(x y+x^2\right) d y=0 \\
& \therefore\left(x y+x^2\right) d y=-y^2 d x \\
& \therefore \frac{d y}{d x}=-\frac{y^2}{x y+x^2} \ldots
\end{aligned}
$
Put $y=t x$...(ii)
Differentiating w.r.t. $x$, we get
$
\frac{d y}{d x}=t+x \frac{d t}{d x}
$
Substituting (ii) and (iii) in (i), we get
$
\begin{aligned}
& \therefore t+x \frac{d t}{d x}=\frac{-t^2 x^2}{x \cdot t x+x^2} \\
& \therefore t+x \frac{d t}{d x}=\frac{-t^2 x^2}{x^2(t+1)} \\
& \therefore x \frac{d t}{d x}=\frac{-t^2}{t+1}-t \\
& \therefore x \frac{d t}{d x}=\frac{-t^2-t^2-t}{t+1} \\
& \therefore x \frac{d t}{d x}=\frac{-\left(2 t^2+t\right)}{t+1} \\
& \therefore \frac{t+1}{2 t^2+t} d t=-\frac{1}{x} d x
\end{aligned}
$
Integrating on both sides, we get
$
\begin{aligned}
& \int \frac{t+1}{2 t^2+t} d t=-\int \frac{1}{x} d x \\
& \therefore \int \frac{2 t+1-t}{t(2 t+1)} d t=-\int \frac{1}{x} d x \\
& \therefore \int \frac{1}{t} d t-\int \frac{1}{2 t+1} d t=-\int \frac{1}{x} d x \\
& \therefore \log |t|-\frac{1}{2} \log |2 t+1|=-\log |x|+\log |c| \\
& \therefore 2 \log | t |-\log |2 t +1|=-2 \log | x |+2 \log | c | \\
& \therefore 2 \log \left|\frac{y}{x}\right|-\log \left|\frac{2 y}{x}+1\right|=-2 \log |x|+2 \log |c| \\
& \therefore 2 \log | y |-2 \log | x |-\log |2 y + x |+\log | x |=-2 \log | x |+2 \log | c | \\
& \therefore \log \left| y ^2\right|+\log | x |=\log \left| c ^2\right|+\log |2 y + x | \\
& \therefore \log \left| y ^2 x \right|=\log \left| c ^2( x +2 y )\right| \\
& \therefore xy ^2= c ^2( x +2 y )
\end{aligned}
$
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Question 85 Marks
The rate of depreciation $\frac{d V}{d t}$ of a machine is inversely proportional to the square of $t+1$, where $V$ is the value of the machine $t$ years after it was purchased. The initial value of the machine was ₹ 8,00,000 and its value decreased $₹ 1,00,000$ in the first year. Find the value after 6 years.
Answer
Let $V$ be the value of the machine at the end of $t$ years. Then $\frac{d V}{d t}$, the rate of depreciation, is inversly proportional to $(t+1)^2$.
$
\therefore \frac{d V}{d t} \propto \frac{1}{(t+1)^2}
$
$\therefore \frac{d V}{d t}=-\frac{k}{(t+1)^2}, k>0$ is a constant
$
\therefore d V=\frac{-k d t}{(t+1)^2}
$
On integrating, we get
$
\begin{aligned}
& \int d V=-k \int \frac{d t}{(t+1)^2} \\
& \therefore V=-k\left[\frac{-1}{t+1}\right]+c \\
& \therefore V=\frac{k}{t+1}+c
\end{aligned}
$
Initially, i.e. when $t =0, V =800000$
$
\therefore 800000=\frac{k}{1}+ C = k + C
$
Now, when $t =1, V =800000-100000=700000$
$
\therefore 700000=\frac{k}{1+1}+ C =\frac{k}{2}+ C
$
Subtracting (2) from (1), we get
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Question 95 Marks
Find the population of a city at any time t, given that the rate of increase of population is proportional to the population at that instant and that in a period of 40 years, the population increased from 30,000 to 40,000.
Answer
Let $P$ be the population of the city at time $t$.
Then $\frac{d P}{d t}$, the rate of increase of population, is proportional to $P$.
$
\therefore \frac{d P}{d t} \propto P
$
$\therefore \frac{d P}{d t}= kP$, where $k$ is a constant.
$
\therefore \frac{d P}{P}= kdt
$
On integrating, we get
$
\begin{aligned}
& \int \frac{1}{P} dP = k \int dt \\
& \therefore \log P = kt + c
\end{aligned}
$
Initially, i.e. when $t=0, P=30000$
$
\therefore \log 30000= k \times 0+ c
$
$
\therefore c =\log 30000
$
$\therefore \log P= kt +\log 30000$
$
\therefore \log P-\log 30000=k t
$
$
\therefore \log \left(\frac{P}{30000}\right)=k t
$
Now, when $t=40, P=40000$
$
\therefore \log \left(\frac{40000}{30000}\right)=k \times 40
$
$\therefore k=\frac{1}{40} \log \left(\frac{4}{3}\right) \quad$
$\therefore$ (1) becomes, $\log \left(\frac{P}{30000}\right)=\frac{t}{40} \log \left(\frac{4}{3}\right)=\log \left(\frac{4}{3}\right)^{\frac{t}{40}}$
$
\begin{aligned}
& \therefore \frac{P}{30000}=\left(\frac{4}{3}\right)^{\frac{t}{40}} \\
& \therefore P=30000\left(\frac{4}{3}\right)^{\frac{t}{40}}
\end{aligned}
$
$\therefore$ the population of the city at time t = 30000 $(\frac{4}{3})^{\frac{t}{40}}$
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Question 105 Marks
The rate of growth of bacteria is proportional to the number present. If initially, there were 1000 bacteria and the number doubles in 1 hour, find the number of bacteria after $\frac{5}{2}$ hours. [Given: √2 = 1.414]
Answer
Let $x$ be the number of bacteria at time $t$.
Then the rate of increase is $\frac{d x}{d t}$ which is proportional to $x$.
$\therefore \frac{d x}{d t} \propto x$
$\therefore \frac{d x}{d t}= kx$, where $k$ is a constant
$
\therefore \frac{d x}{x}= kdt
$
On integrating, we get
$\int \frac{d x}{x}= k \int dt$
$
\therefore \log x = kt + c
$
Initially, i.e. when $t=0, x=1000$
$\therefore \log 1000= k \times 0+ c$
$\therefore c =\log 1000$
$\therefore \log x=k t+\log 1000$
$\therefore \log x -\log 1000= kt$
$
\therefore \log \left(\frac{x}{1000}\right)= kt
$
Now, when $t =1, x =2 \times 1000=2000$
$
\therefore \log \left(\frac{2000}{1000}\right)=k \quad \therefore k=\log 2
$
$\therefore$ (1) becomes, $\log \left(\frac{x}{1000}\right)=t \log 2$
If $t=\frac{5}{2}$, then
$
\begin{aligned}
& \log \left(\frac{x}{1000}\right)=\frac{5}{2} \log 2=\log (2)^{\frac{5}{2}} \\
\therefore & \left(\frac{x}{1000}\right)=(2)^{\frac{5}{2}}=4 \sqrt{2}=4 \times 1.414=5.656 \\
\therefore & x=5.656 \times 1000=5656
\end{aligned}
$
$\therefore$ number of bacteria after $\frac{5}{2}$ hours $=5656$.
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Question 115 Marks
If the population of a town increases at a rate proportional to the population at that time. If the population increases from 40 thousand to 60 thousand in 40 years, what will be the population in another 20 years?(Given: $\sqrt{\frac{3}{2}}=1.2247$ )
Answer
Let $P$ be the population of the city at time $t$. Then $\frac{d P}{d t}$, the rate of increase of population, is proportional to $P$.
$
\begin{aligned}
& \therefore \frac{d P}{d t} \propto P \\
& \therefore \frac{d P}{d t}= kP , k \text { is a constant } \\
& \therefore \frac{d P}{P}= kdt
\end{aligned}
$
Integrating, we get
$
\begin{aligned}
& \int \frac{d P}{P}= k \int dt \\
& \therefore \log P = kt + c
\end{aligned}
$
Initially, i.e. when $t=0, P=40000$
$
\begin{aligned}
& \therefore \log 40000=0+c \\
& \therefore c=\log 40000 \\
& \therefore \log P=k t+\log 40000 \\
& \therefore \log P-\log 40000=k t \\
& \therefore \log \left(\frac{P}{40000}\right)= kt \ldots \ldots \ldots . .(1)
\end{aligned}
$
When $t =40, P =60000$
$
\therefore \log \left(\frac{60000}{40000}\right)=40 k
$
$\therefore k=\frac{1}{40} \log \left(\frac{3}{2}\right)$
$\therefore$ (1) becomes
$
\log \left(\frac{P}{40000}\right)=\frac{t}{40} \log \left(\frac{3}{2}\right)
$
$
\begin{gathered}
=\log \left(\frac{3}{2}\right)^{\frac{t}{40}} \\
\therefore \frac{P}{40000}=\left(\frac{3}{2}\right)^{\frac{t}{40}}
\end{gathered}
$
We have to find $P$ in another 20 years i.e. at $t=40+20=60$
If $t=60$, then
$
\begin{aligned}
& \frac{P}{40000}=\left(\frac{3}{2}\right)^{\frac{60}{40}}=\left(\frac{3}{2}\right)^{\frac{3}{2}}=\frac{3}{2} \sqrt{\frac{3}{2}} \\
& \therefore P=\frac{40000 \times 3}{2} \times 1.2247 \\
& =73482
\end{aligned}
$
... [By data]
$\therefore$ population after 60 years will be 73482 .
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Question 125 Marks
In a certain culture of bacteria, the rate of increase is proportional to the number present. If it is found that the number doubles in 4 hours, find the number of times the bacteria are increased in 12 hours.
Answer
Let $x$ be the number of bacteria in the culture at time $t$. Then the rate of increase is $\frac{d x}{d t}$ which is proportional to $x$. $\therefore \frac{d x}{d t} \propto x$
$\therefore \frac{d x}{d t}= kx$, where $k$ is a constant
$
\therefore \frac{d x}{x}= kdt
$
On integrating, we get
$
\begin{aligned}
& \int \frac{d x}{x}= k \int dt \\
& \therefore \log x = kt + c
\end{aligned}
$
Initially, i.e. when $t=0$, let $x=x_0$
$
\begin{aligned}
& \therefore \log x _0= k \times 0+ c \\
& \therefore c =\log x _0 \\
& \therefore \log x = kt +\log x _0 \\
& \therefore \log x -\log x _0= kt \\
& \therefore \log \left(\frac{x}{x_0}\right)= kt . . . . .(1)
\end{aligned}
$
Since the number doubles in 4 hours, i.e. when $t=4, x=$ $2 x_0$
$
\therefore \log \left(\frac{2 x_0}{x_0}\right)=4 k \quad \therefore k=\frac{1}{4} \log 2
$
$\therefore$ (1) becomes, $\log \left(\frac{x}{x_0}\right)=\frac{t}{4} \log 2$
When $t=12$, we get
$
\begin{gathered}
\log \left(\frac{x}{x_0}\right)=\frac{12}{4} \log 2=3 \log 2 \\
\therefore \log \left(\frac{x}{x_0}\right)=\log 8 \\
\therefore \frac{x}{x_0}=8 \quad \therefore x=8 x_0
\end{gathered}
$
$\therefore$ the number of bacteria will be 8 times the original number in 12 hours.
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Question 135 Marks
Solve the following differential equations : $(x+y) \frac{d y}{d x}=1$
Answer
$
\begin{aligned}
& ( x + y ) \frac{d y}{d x}=1 \\
& \therefore \frac{d x}{d y}= x + y \\
& \therefore \frac{d x}{d y}- x = y \\
& \therefore \frac{d x}{d y}+(-1) x = y
\end{aligned}
$
This is the linear differential equation of the form
$
\begin{aligned}
& \frac{d x}{d y}+P \cdot x=Q, \text { where } P=-1 \text { and } Q=y \\
& \therefore \text { I.F. }=e^{\int P d y}=e^{\int-1 d y}=e^{-y}
\end{aligned}
$
$\therefore$ the solution of (1) is given by
$
\begin{array}{rl}
& x \cdot(\text { I.F. })=\int Q \cdot(\text { I.F. }) d y+c \\
\therefore x \cdot e^{-y} & =\int y \cdot e^{-y} d y+c \\
\therefore & e^{-y} \cdot x=y \int e^{-y} d y-\int\left[\frac{d}{d y}(y) \int e^{-y} d y\right] d y+c \\
= & y \cdot \frac{e^{-y}}{-1}-\int 1 \cdot \frac{e^{-y}}{-1} d y+c \\
& =-y e^{-y}+\int e^{-y} d y+c \\
\therefore & e^{-y} \cdot x=-y e^{-y}+\frac{e^{-y}}{-1}+c \\
\therefore & e^{-y} \cdot x+y e^{-y}+e^{-y}=c \\
\therefore & e^{-y}(x+y+1)=c \\
\therefore x+y & x=c e^y
\end{array}
$
This is the general solution.
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Question 145 Marks
Solve the following differential equations : $x \frac{d y}{d x}+2 y=x^2 \cdot \log x$
Answer
$
\begin{aligned}
& x \frac{d y}{d x}+2 y = x ^2 \cdot \log x \\
& \therefore \frac{d y}{d x}+\left(\frac{2}{x}\right) \cdot y=x \cdot \log x
\end{aligned}
$
This is the linear differential equation of the form
$
\begin{aligned}
& \frac{d y}{d x}+P \cdot y=Q \text {, where } P=\frac{2}{x} \text { and } Q=x \cdot \log x \\
& \therefore \text { I.F. }=e^{\int P d x}=e^{\int \frac{2}{x} d x}=e^{2 \int \frac{1}{x} d x} \\
& =e^{2 \log x}=e^{\log x^2}=x^2
\end{aligned}
$
$\therefore$ the solution of (1) is given by
$
\begin{aligned}
y \cdot(\text { I.F. }) & =\int Q \cdot \text { (I.F.) } d x+c \\
\therefore y \cdot x^2 & =\int(x \log x) \cdot x^2 d x+c \\
\therefore x^2 \cdot y & =\int x^3 \cdot \log x d x+c \\
= & (\log x) \int x^3 d x-\int\left[\frac{d}{d x}(\log x) \int x^3 d x\right] d x+c \\
= & (\log x) \cdot \frac{x^4}{4}-\int \frac{1}{x} \cdot \frac{x^4}{4} d x+c \\
= & \frac{1}{4} x^4 \log x-\frac{1}{4} \int x^3 d x+c
\end{aligned}
$
$
\begin{aligned}
& \therefore x^2 \cdot y=\frac{1}{4} x^4 \log x-\frac{1}{4} \cdot \frac{x^4}{4}+c \\
& \therefore y \cdot x^2=\frac{x^4 \log x}{4}-\frac{x^4}{16}+c
\end{aligned}
$
This is the general solution.
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Question 155 Marks
Solve the following differential equations : $x ^2 \frac{d y}{d x}= x ^2+ xy - y ^2$
Answer
$
\begin{aligned}
& x^2 \frac{d y}{d x}=x^2+x y-y^2 \\
& \therefore \frac{d y}{d x}=1+\frac{y}{x}-\frac{y^2}{x^2}
\end{aligned}
$
Put $y=v x \quad$ i.e. $\frac{y}{x}=v$
$
\therefore \frac{d y}{d x}=v+x \frac{d v}{d x}
$
$\therefore$ (1) becomes
$
\begin{aligned}
& v+x \frac{d v}{d x}=1+v-v^2 \\
& \therefore x \frac{d v}{d x}=1-v^2 \\
& \therefore \frac{d v}{1-v^2}=\frac{d x}{x}
\end{aligned}
$
Integrating, we get
$
\begin{aligned}
& \int \frac{d v}{1-v^2}=\int \frac{d x}{x} \\
& \therefore \frac{1}{2} \log \left|\frac{1+v}{1-v}\right|=\log x+\log c_1
\end{aligned}
$
$
\begin{aligned}
\therefore \frac{1}{2} \log \left|\frac{1+\frac{y}{x}}{1-\frac{y}{x}}\right|=\log \left(x c_1\right) & \\
\therefore \log \left|\frac{x+y}{x-y}\right| & =2 \log \left(x c_1\right) \\
& =\log \left(x^2 c_1{ }^2\right)
\end{aligned}
$
$\therefore \frac{x+y}{x-y}=c x^2, c=c_1{ }^2$, is the required solution.
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Question 165 Marks
Solve the following differential equations : $xy \frac{d y}{d x}= x ^2+2 y ^2$
Answer
$
\begin{array}{r}
x y \frac{d y}{d x}=x^2+2 y^2 \\
\therefore \frac{d y}{d x}=\frac{x^2+2 y^2}{x y}
\end{array}
$
Put $y=v x$. Then $\frac{d y}{d x}=v+x \frac{d v}{d x}$
$\therefore$ (1) becomes, $v+x \frac{d v}{d x}=\frac{x^2+2 v^2 x^2}{x \cdot v x}=\frac{1+2 v^2}{v}$
$
\therefore x \frac{d v}{d x}=\frac{1+2 v^2}{v}-v=\frac{1+2 v^2-v^2}{v}
$
$\therefore x \frac{d v}{d x}=1+v^2$
$\therefore x \frac{d v}{d x}=\frac{1+v^2}{v}$
$
\therefore \frac{v}{1+v^2} d v=\frac{1}{x} d x
$
Integrating, we get
$
\begin{aligned}
& \int \frac{v}{1+v^2} d v=\int \frac{1}{x} d x \\
\therefore & \frac{1}{2} \int \frac{2 v}{1+v^2} d v=\int \frac{1}{x} d x+\log c_1 \\
\therefore & \frac{1}{2} \log \left|1+v^2\right|=\log |x|+\log c_1 \\
\therefore & \log \left|1+v^2\right|=2 \log |x|+2 \log c_1 \\
\therefore & \log \left|1+v^2\right|=\log \left|x^2\right|+\log c_1{ }^2 \\
\therefore & \log \left|1+v^2\right|=\log \left|c x^2\right|, \text { where } c=c_1{ }^2 \\
\therefore & 1+v^2=c x^2
\end{aligned}
$
$
\begin{aligned}
& \therefore 1+\frac{y^2}{x^2}=c x^2 \\
& \therefore \frac{x^2+y^2}{x^2}=c x^2 \\
& \therefore x^2+y^2=c x^4
\end{aligned}
$
This is the general solution.
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Question 175 Marks
Solve the following differential equations : $\left(x^2-y^2\right) d x+2 x y d y=0$
Answer
$
\begin{aligned}
& \left(x^2-y^2\right) d x+2 x y d y=0 \\
& \therefore 2 x y d y=-\left(x^2-y^2\right) d x=\left(y^2-x^2\right) d x
\end{aligned}
$
$
\therefore \frac{d y}{d x}=\frac{y^2-x^2}{2 x y}
$
Put $y=v x \quad \therefore \frac{d y}{d x}=v+x \frac{d v}{d x}$
$\therefore$ (1) becomes, $v+x \frac{d v}{d x}=\frac{v^2 x^2-x^2}{2 x \cdot v x}$
$\therefore v+x \frac{d v}{d x}=\frac{v^2-1}{2 v}$
$\therefore x \frac{d v}{d x}=\frac{v^2-1}{2 v}-v=\frac{v^2-1-2 v^2}{2 v}$
$\therefore x \frac{d v}{d x}=\frac{-1-v^2}{2 v}=-\left(\frac{1+v^2}{2 v}\right)$
$
\therefore \frac{2 v}{1+v^2} d v=-\frac{1}{x} d x
$
Integrating, we get
$
\begin{aligned}
& \int \frac{2 v}{1+v^2} d v=-\int \frac{1}{x} d x \\
\therefore & \log \left|1+v^2\right|=-\log x+\log c \\
\therefore & {\left[\because \frac{d}{d v}\left(1+v^2\right)=2 v \text { and } \int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|+c\right] } \\
\therefore & \log \left|1+\frac{y^2}{x^2}\right|=-\log x+\log c \\
\therefore & \log \left|\frac{x^2+y^2}{}\right|=\log \left|\frac{c}{-c}\right|
\end{aligned}
$
$
\begin{aligned}
& \therefore \frac{x^2+y^2}{x^2}=\frac{c}{x} \\
& \therefore x^2+y^2=c x
\end{aligned}
$
This is the general solution.
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Question 185 Marks
Solve the following differential equations : $\frac{d y}{d x}+\frac{x-2 y}{2 x-y}=0$
Answer
$
\begin{array}{r}
\frac{d y}{d x}+\frac{x-2 y}{2 x-y}=0 \\
\therefore \frac{d y}{d x}=-\left(\frac{x-2 y}{2 x-y}\right)
\end{array}
$
Put $y=v x . \quad \therefore \frac{d y}{d x}=v+x \frac{d v}{d x}$
$\therefore$ (1) becomes, $v+x \frac{d v}{d x}=-\left(\frac{x-2 v x}{2 x-v x}\right)$
$
\begin{aligned}
& \therefore v+x \frac{d v}{d x}=-\left(\frac{1-2 v}{2-v}\right) \\
& \therefore x \frac{d v}{d x}=-\left(\frac{1-2 v}{2-v}\right)-v \\
& \therefore x \frac{d v}{d x}=\frac{-1+2 v-2 v+v^2}{2-v} \\
& \therefore x \frac{d v}{d x}=\frac{v^2-1}{2-v} \\
& \therefore \frac{2-v}{v^2-1} d v=\frac{1}{x} d x
\end{aligned}
$
Integrating, we get
$
\begin{aligned}
& \int \frac{2-v}{v^2-1} d v=\int \frac{1}{x} d x \\
& \therefore 2 \int \frac{1}{v^2-1} d v-\frac{1}{2} \int \frac{2 v}{v^2-1} d v=\int \frac{1}{x} d x \\
& \therefore 2 \times \frac{1}{2} \log \left|\frac{v-1}{v+1}\right|-\frac{1}{2} \log \left|v^2-1\right|=\log |x|+\log c_1
\end{aligned}
$
$
\begin{aligned}
& \cdots\left[\because \frac{d}{d v}\left(v^2-1\right)=2 v \text { and } \int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|+c\right] \\
\therefore & \log \left|\frac{v-1}{v+1}\right|-\log \left|\left(v^2-1\right)^{\frac{1}{2}}\right|=\log \left|c_1 x\right| \\
\therefore & \log \left|\frac{v-1}{v+1} \cdot \frac{1}{\sqrt{v^2-1}}\right|=\log \left|c_1 x\right| \\
\therefore & \frac{v-1}{v+1} \cdot \frac{1}{\sqrt{v^2-1}}=c_1 x \\
\therefore & \frac{y}{x-1} \cdot \frac{1}{\frac{y}{x}+1}=c_1 x \\
\therefore & \frac{y-x}{y+x} \cdot \frac{x}{\sqrt{x^2}-1} \\
\therefore & \frac{y-x}{y+x}=c_1 x \\
\therefore & \frac{y-x}{y+x}=c_1 \sqrt{y-x} \cdot \sqrt{y+x} \\
\therefore & \sqrt{y-x}=c_1(y+x)^{\frac{3}{2}} \\
\therefore & y-x=c_1^2(x+y)^3 \\
\therefore & y-x=c(x+y)^3, \text { where } c=c_1{ }^2
\end{aligned}
$
This is the general solution.
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Question 195 Marks
Solve the following differential equations : $x^2 y d x-\left(x^3+y^3\right) d y=0$
Answer
$
\begin{aligned}
& x ^2 y dx -\left( x ^3+ y ^3\right) d y=0 \\
& \therefore\left( x ^3+ y ^3\right) d y= x ^2 y dx \\
& \therefore \frac{d y}{d x}=\frac{x^2 y}{x^3+y^3} \ldots \ldots .(1)
\end{aligned}
$
Put $y = vx$
$
\therefore \frac{d y}{d x}=v+x \frac{d v}{d x}
$
$\therefore$ (1) becomes, $v+x \frac{d v}{d x}=\frac{x^2 \cdot v x}{x^3+v^3 x^3}=\frac{v}{1+v^3}$
$
\therefore x \frac{d v}{d x}=\frac{v}{1+v^3}-v=\frac{v-v-v^4}{1+v^3}
$
$\therefore x \frac{d v}{d x}=\frac{-v^4}{1+v^3}$
$
\therefore \frac{1+v^3}{v^4} d v=-\frac{1}{x} d x
$
Integrating, we get
$
\begin{aligned}
& \int \frac{1+v^3}{v^4} d v=-\int \frac{1}{x} d x \\
\therefore & \int\left(\frac{1}{v^4}+\frac{1}{v}\right) d v=-\int \frac{1}{x} d x \\
\therefore & \int v^{-4} d v+\int \frac{1}{v} d v=-\int \frac{1}{x} d x \\
\therefore & \frac{v^{-3}}{-3}+\log |v|=-\log |x|+c_1 \\
\therefore & -\frac{1}{3 v^3}+\log |v|=-\log |x|+c_1
\end{aligned}
$
$
\begin{aligned}
& \therefore-\frac{1}{3} \cdot \frac{1}{\left(\frac{y}{x}\right)^3}+\log \left|\frac{y}{x}\right|=-\log |x|+c_1 \\
& \therefore-\frac{x^3}{3 y^3}+\log |y|-\log |x|=-\log |x|-\log c \\
& \text{where }c_1=-\log c, \\
& \therefore \frac{x^3}{3 y^3}=\log c+\log y \\
& \therefore \frac{x^3}{3 y^3}=\log |c y|
\end{aligned}
$
This is the general solution.
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Question 215 Marks
In each of the following examples, verify that the given function is a solution of the corresponding differential equation:
SolutionD.E.
(i)xy=log y+ky^(')(1-xy)=y^(2)
(ii)y=x^(n)x^(2)(d^(2)y)/(dx^(2))-nx(dy)/(dx)+ny=0
(iii)y=e^(x)(dy)/(dx)=y
(iv)y=1-log xx^(2)(d^(2)y)/(dx^(2))=1
(v)y=ae^(x)+be^(-x)(d^(2)y)/(dx^(2))=y
(vi)ax^(2)+by^(2)=5xy(d^(2)y)/(dx^(2))+x((dy)/(dx))^(2)=y*(dy)/(dx)
Answer
(i) $x y=\log y+k$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
& x \cdot \frac{d y}{d x}+y \times 1=\frac{1}{y} \cdot \frac{d y}{d x}+0 \\
& \therefore x \frac{d y}{d x}+y=\frac{1}{y} \cdot \frac{d y}{d x} \\
& \left(x-\frac{1}{y}\right) \frac{d y}{d x}=-y
\end{aligned}
$
$\therefore\left(\frac{x y-1}{y}\right) \frac{d y}{d x}=-y$
$\therefore \frac{d y}{d x}=\frac{-y^2}{x y-1}=\frac{y^2}{1-x y}$, if $x y \neq 1$
$\therefore y^{\prime}=\frac{y^2}{1-x y^{\prime}}$, if $x y \neq 1$.
$\therefore y^{\prime}(1-x y)=y^2$
Hence, $x y=\log y+k$ is a solution of the D.E. $y^{\prime}(1-x y)=y^2$.
(ii) $y=x^n$
Differentiating twice w.r.t. $x$, we get
$
\begin{aligned}
& \frac{d y}{d x}=\frac{d}{d x}\left(x^n\right)=n x^{n-1} \\
& \text { and } \frac{d^2 y}{d x^2}=\frac{d}{d x}\left(n x^{n-1}\right)=n \frac{d}{d x}\left(x^{n-1}\right)=n(n-1) x^{n-2} \\
& \begin{aligned}
\therefore x^2 \frac{d^2 y}{d x^2}-n x \frac{d y}{d x}+n y \\
& =x^2 \cdot n(n-1) x^{n-2}-n x \cdot n x^{n-1}+n \cdot x^n \\
& =n(n-1) x^n-n^2 x^n+n x^n \\
& =\left(n^2-n-n^2+n\right) x^n=0
\end{aligned}
\end{aligned}
$
This shows that $y=x^n$ is a solution of the D.E.
$
x^2 \frac{d^2 y}{d x^2}-n x \frac{d y}{d x}+n y=0
$
(iii) $y = e ^{ x }$
Differentiating w.r.t. $x$, we get
$
\frac{d y}{d x}= e ^{ x }= y
$
Hence, $y = ex$ is a solution of the D.E. $\frac{d y}{d x}= y$.
(iv) $y=1-\log x$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
\frac{d y}{d x} & =\frac{d}{d x}(1)-\frac{d}{d x}(\log x) \\
& =0-\frac{1}{x}=-\frac{1}{x}
\end{aligned}
$
Differentiating again w.r.t. $x$, we get
$
\begin{aligned}
& \frac{d^2 y}{d x^2}=-\frac{d}{d x}\left(x^{-1}\right)=-(-1) x^{-2}=\frac{1}{x^2} \\
& \therefore x^2 \frac{d^2 y}{d x^2}=1
\end{aligned}
$
Hence, $y=1-\log x$ is a solution of the D.E.
$
x^2 \frac{d^2 y}{d x^2}=1
$
(v) $y=a e^x+b e^{-x}$
Differentiating w.r.t. $x$, we get
$
\frac{d y}{d x}= a \left( e ^{ x }\right)+ b \left(- e ^{- x }\right)= ae ^{ x }- be ^{- x }
$
Differentiating again w.r.t. $x$, we get
$
\begin{aligned}
& \frac{d^2 y}{d x^2}= a \left( e ^{ x }\right)- b \left(- e ^{- x }\right) \\
& = ae ^{ x }+ be ^{- x }
\end{aligned}
$
$=y$
Hence, $y = ae ^{ x }+ be ^{- x }$ is a solution of the D.E. $\frac{d^2 y}{d x^2}= y$.
(vi) $a x^2+b y^2=5$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
& a(2 x)+b\left(2 y \frac{d y}{d x}\right)=0 \\
& \therefore a x+b y \frac{d y}{d x}=0
\end{aligned}
$
$
\therefore a x=-b y \frac{d y}{d x}
$
Differentiating again w.r.t. $x$, we get
$
\begin{aligned}
& a \cdot 1=-b\left[y \frac{d}{d x}\left(\frac{d y}{d x}\right)+\frac{d y}{d x} \cdot \frac{d y}{d x}\right] \\
& \therefore a=-b\left[y \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2\right]
\end{aligned}
$
Dividing (1) by (2), we get
$
\begin{aligned}
& x=\frac{y \frac{d y}{d x}}{y\left(\frac{d^2 y}{d x^2}\right)+\left(\frac{d y}{d x}\right)^2} \\
& \therefore x y \frac{d^2 y}{d x^2}+x\left(\frac{d y}{d x}\right)^2=y \frac{d y}{d x}
\end{aligned}
$
Hence, $a x^2+b y^2=5$ is a solution of the D.E.
$
x y \frac{d^2 y}{d x^2}+x\left(\frac{d y}{d x}\right)^2=y\left(\frac{d y}{d x}\right)
$
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