3 Marks Question

Question 13 Marks

Doctors have shown certain drugs leave a person's bloodstream at a rate that is proportional to the amount present. In an experiment a patient is injected with 450 mg of a substance. Seven hours later it is found that 50 mg of the substance remains. Assuming the proportional model is correct for the particular substance.
(a) Express the differential equation that models this scenario.
(b) Find the time constant, k.(log 9 = 2.197225).

Answer

(a) With the amount of the substance represented as A(t) we can write the differential equation as $\frac{d A}{d t}=k A$
(b) To find the time constant, k, we first write the expression for the general solution as $A(t)=A(0) e^{k t}=450 e^{k t}$
Where, A(0) = 450
Next, we use the fact that 50 mg was found in the bloodstream after 7 hours to solve for k.
$\begin{aligned} A(7) & =450 e^{k .7} \\ 50 & =450 e^{k .7} \\ \frac{1}{9} & =e^{k .7} \\ \log \frac{1}{9} & =k .7\end{aligned}$
$\log \frac{1}{9}=k=\frac{-\log 9}{7} \cong-0.31$
As expected, k is negative since the substance decays over time.

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Question 23 Marks

The surface area of a balloon being inflated changes at a constant rate. If initially, its radius is 3 units and after 2 seconds, it is 5 units, find the radius after t seconds.

Answer

Let r be the radius and S be the surface area of the balloon at any time t. Then,
$S=4 \pi r^2$
$\Rightarrow \quad \frac{d S}{d t}=8 \pi r \frac{d r}{d t}$ ...(i)
It is given that, $\frac{d S}{d t}=$ constant $=k$ (say $)$
Putting $\frac{d S}{d t}$ in eq. (i), we get
$k=8 \pi r \frac{d r}{d t}$
$\Rightarrow \quad k=8 \pi r \frac{d r}{d t}$
$\Rightarrow \quad 8 \pi r d r=k d t$
(By separating the variables)
Integrating both sides, we get
$4 \pi r^2=k t+C$ ....(ii)
Given that at t = 0 r = 3 Putting t = 0 andr r = 3 in eq. (ii), we get
$36 \pi=k(0)+C$
$\Rightarrow$ $C=36 \pi$
Putting t = 2 and r = 5 in eq. (ii), we get
$100 \pi=2 k+36 \pi$
$\Rightarrow$ $k=32 \pi$
Substituting the values of C and k in eq. (ii), we get
$4 \pi r^2=32 \pi t+36 \pi$
$\Rightarrow \quad r^2=8 t+9$
$\Rightarrow \quad r=\sqrt{8 t+9}$

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Question 33 Marks

The population of a town grows at the rate of 10% per year. Using differential equation, find how long will it take for the population to grow 4 times.

Answer

Let initial population be: $P_0$ and population after t years be P
$\frac{d P}{d t}=10 \%$ of $P$
$\Rightarrow \quad \frac{d P}{d t}=\frac{1}{10} P$
$\Rightarrow \quad \frac{d P}{P}=\frac{1}{10} d t$
Integrating both sides, we get
$\int \frac{d P}{P}=\int \frac{1}{10} d t$
$\log P=\frac{1}{10} t+C$
When $t=0, P=P_0 \Rightarrow \log P_0=C$
$\log P=\frac{1}{10} \cdot t+\log P_0$
$\log \left(\frac{P}{P_0}\right)=\frac{1}{10} . t$
Since, population grows 4 times, $P=4 P_0$
$\therefore \quad \log 4=\frac{1}{10} t$
$\Rightarrow \quad t=10 \log 4$
$\Rightarrow 20 \log 2$ (vears)

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Question 43 Marks

Solve the differential equation: $\frac{d y}{d x}=e^{x+y}+x^2 e^y$

Answer

Given, $\quad \frac{d y}{d x}=e^{x+y}+x^2 e^y$
$=e^x e^y+x^2 e^y$
$\Rightarrow \quad \frac{d y}{d x}=e^y\left(e^x+x^2\right)$
$\int \frac{d y}{e^y}=\int\left(e^x+x^2\right) d x$
(integrating both sides)
$\Rightarrow \quad \int e^{-y} d y=\int e^x d x+\int x^2 d x$
$\Rightarrow \quad-e^{-y}=e^x+\frac{x^3}{3}+c$
$\Rightarrow \quad-\frac{1}{e^y}=e^x+\frac{x^3}{3}+c$
or $\quad e^{x+y}+\frac{e^y \cdot x^3}{3}+c e^y+1=0$
is the required solution

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Question 53 Marks

Solve the differential equation: $(x+1) d y-2 x y d x=0$

Answer

$(x+1) d y-2 x y d x=0$
$\Rightarrow \quad \int \frac{d y}{y}=\int \frac{2 x}{x+1} d x$
$\Rightarrow \quad \log y=2 \int \frac{x+1-1}{x+1} d x$
$=2 \int\left(1-\frac{1}{x+1}\right) d x$
$\Rightarrow \log y=2[x-\log (x+1)]+\log c$
$\log y+2 \log (x+1)-\log c=2 x$
$\Rightarrow \quad \log \frac{y(x+1)^2}{c}=2 x$
$\Rightarrow \quad \frac{y(x+1)^2}{c}=e^{2 x}$
$\Rightarrow \quad y=\frac{c e^{2 x}}{(x+1)^2}$

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Question 63 Marks

Solve : $\left(x^2-y x^2\right) d y+\left(y^2+x y^2\right) d x=0$

Answer

Given, $\left(x^2-y x^2\right) d y+\left(y^2+x y^2\right) d x=0$
$\Rightarrow \quad x^2(1-y) d y+y^2(1+x) d x=0$
$\Rightarrow \quad \frac{1-y}{y^2} d y+\frac{1+x}{x^2} d x=0$
On integration, we get
$\int \frac{1-y}{y^2} d y+\int \frac{1+x}{x^2} d x=0$
$\Rightarrow \int \frac{1}{y^2} d y-\int \frac{1}{y} d y+\int \frac{1}{x^2} d x+\int \frac{1}{x} d x=0$
$\Rightarrow \quad \log \left(\frac{x}{y}\right)-\frac{1}{x}-\frac{1}{y}= C$
$\Rightarrow \quad \log \left(\frac{x}{y}\right)=C+\frac{1}{x}+\frac{1}{y}$

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Question 73 Marks

Find the particular solution of the following differential equation: $x y \frac{d y}{d x}=(x+2)(y+2) ; y=-1$, when $x=1$.

Answer

$x y \frac{d y}{d x}=(x+2)(y+2)$
or, $\frac{y}{y+2} d y=\frac{x+2}{x} d x$
or $\left(1-\frac{2}{y+2}\right) d y=\left(1+\frac{2}{x}\right) d x$
On integrating it, we get
$y-2 \log (y+2)=x+2 \log x+C$
Given $y=-1$, when $x=1$, then from $(i$)
$-1-2 \log (-1+2)=1+2 \log 1+C$
$-1-2 \log (-1+2)=1+2 \log 1+C$
or, $C=-2$, as $\log 1=0$
Then (i) becomes:
$\begin{array}{l}y-2 \log (y+2)=x+2 \log x-2 \\ \text { or } y=x-2+2\{\log (y+2)+\log x\}\end{array}$
or $y=x-2+2 \log \{x(y+2)\}$
This is the required particular solution.

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Question 83 Marks

Find the particular solution of the following differential equation: $(x+1) \frac{d y}{d x}=2 e^{-y}-1 ; y=0$ when $x=0$.

Answer

Given equation can be written as
$\int \frac{d y}{2 e^{-y}-1}=\int \frac{d x}{x+1}$
$\Rightarrow \int \frac{e^y}{2-e^y} d y=\int \frac{d x}{x+1}$
$\Rightarrow-\log \left|2-e^y\right|+\log c=\log |x+1|$
$\Rightarrow \quad\left(2-e^y\right)(x+1)=c$
When $x=0, y=0 \Rightarrow c=1$
$\therefore$ Solution is $\left(2-e^y\right)(x+1)=1$

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Question 93 Marks

Find the particular solution of the differential equation $\log \left(\frac{d y}{d x}\right)=3 x+4 y$, given that $y=0$ when x = 0

Answer

Given differential equation can be written as
$\frac{d y}{d x}=e^{(3 x+4 y)}=e^{3 x} \cdot e^{4 y}$
$\therefore \quad \int e^{-4 y} d y=\int e^{3 x} d x$
or, $\frac{e^{-4 y}}{-4}=\frac{e^{3 x}}{3}+C$
$\therefore \quad 4 e^{3 x}+3 e^{-4 y}+12 C=0$
Taking x = 0 y = 0 we get
$C=-\frac{7}{12}$
$\therefore$ The solution is $4 e^{3 x}+3 e^{-t y}-7=0$.

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Question 103 Marks

Find the particular solution of the differential equation $\frac{d y}{d x}=1+x+y+x y$ given that $y=0$ when x = 1

Answer

$\frac{d y}{d x}=1+x+y+x y$
= (1 + x)(1 + y)
or, $\int \frac{d y}{1+y}=\int(1+x) d x$
$\therefore \quad \log |1+y|=x+\frac{x^2}{2}+C$
When, $x=1, y=0$ or $C=-\frac{3}{2}$
$\therefore$ Solution is:
$\log |1+y|=x+\frac{x^2}{2}-\frac{3}{2}$

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Question 113 Marks

Find the particular solution of the differential equation $e^x \sqrt{1-y^2} d x+\left(\frac{y}{x}\right) d y=0$ given that y = 1 when x = 0

Answer

$e^x \sqrt{1-y^2} d x=\frac{-y}{x} d y$
or, $x e^x d x=\frac{-y}{\sqrt{1-y^2}} d y$
Integrating both sides, we get
$\int x e^x d x=\frac{1}{2} \int \frac{-2 y}{\sqrt{1-y^2}} d y$
or, $x e^x-e^x=\sqrt{1-y^2}+C$
For x = 0, y = 1, C = - 1
$\therefore$ Solution is: $e^x(x-1)=\sqrt{1-y^2}-1$

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Question 123 Marks

Prove that $x^2-y^2=C\left(x^2+y^2\right)^2$ is the general solution of diffrential equations $\left(x^2-3 x y^2\right) d x$ $=\left(y^3-3 x^2 y\right) d y$, where C is a parameter.

Answer

$x^2-y^2=C\left(x^2+y^2\right)^2$
or $2 x-2 y y^{\prime}=2 C \left(x^2+y^2\right)\left(2 x+2 y y^{\prime}\right)$
or $\left(x-y y^{\prime}\right)=\frac{x^2-y^2}{y^2+x^2}\left(2 x+2 y y^{\prime}\right)$
or $\left(y^2+x^2\right)\left(x-y y^{\prime}\right)=\left(x^2-y^2\right)\left(2 x+2 y y^{\prime}\right)$
or $\left[-2 y\left(x^2-y^2\right)-y\left(y^2+x^2\right)\right] \frac{d y}{d x}$
$=2 x\left(x^2-y^2\right)-x\left(y^2+x^2\right)$
or $\left(y^3-3 x^2 y\right) \frac{d y}{d x}=\left(x^3-3 x y^2\right)$
or $\left(y^3-3 x^2 y\right) d y=\left(x^3-3 x y^2\right) d x$
Hence $x^2-y^2=C\left(x^2+y^2\right)$ is the solution of given differential equation.

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Question 133 Marks

Form the differential equation of the family of circles in the second quadrant and touching the co-ordinate axes.

Answer

Equation of family of circle is
$(x+a)^2+(y-a)^2=a^2$
or $x^2+y^2+2 a x-2 a y+a^2=0$ ....(i)
Differentiating, we get
$2 x+2 y \frac{d y}{d x}+2 a-2 a \frac{d y}{d x}=0$

or, $x+y \frac{d y}{d x}=a\left(\frac{d y}{d x}-1\right)$
or, $a=\frac{x+y y^{\prime}}{y^{\prime}-1}$,
where $y^{\prime}=\frac{d y}{d x}$
Substituting the value of a in eqn. (i) and simplifying
$\left(x y^{\prime}-x+x+y y^{\prime}\right)^2+\left(y y^{\prime}-y-x-y y^{\prime}\right)^2$
$=\left(x+y y^{\prime}\right)^2$
or, $(x+y)^2\left[\left(y^{\prime}\right)^2+1\right]=\left(x+y y^{\prime}\right)^2$

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Applied Maths 3 Marks Question

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