Chapter 9: Differential Equations

Ans: Let the principal be $\text{p}$, according to  principle increases at the rate of $\text{5}$% per year.

$\frac{dp}{dt}=\left( \frac{5}{100} \right)p$ 

Simplify the expression:

$\frac{dp}{dt}=\frac{p}{20}$ 

$\Rightarrow \frac{dp}{p}=\frac{1}{20}dt$ 

Integrate both side:

$\int \frac{dp}{p}=\int \frac{1}{20}dt$ 

$\Rightarrow \log p=\frac{t}{20}+C$ 

$\Rightarrow p={{e}^{\frac{t}{20}+C}}$ 

$\Rightarrow p=A{{e}^{\frac{t}{20}}}\,\,\left( A={{e}^{C}} \right)$
At $\text{t=0}$, $\text{p=1000}$: 

$1000=A{{e}^{\frac{0}{20}}}$ 

$\Rightarrow A=1000$ 

Thus the relation of principal and time relation:

$\Rightarrow p=1000{{e}^{\frac{t}{20}}}$ 

At $\text{t=10}$: 

$p=1000{{e}^{\frac{10}{20}}}$ 

$\Rightarrow p=1000{{e}^{0.5}}$

$\Rightarrow p=1000\times 1.648$ 

$\Rightarrow p=1648$  

Thus after $\text{10}$ this year the amount will become Rs. $\text{1648}$.

Ans: Let the number of bacteria be $\text{y}$ at time $\text{t}$. According to :

$\frac{dy}{dt}\propto y$ 

$\frac{dy}{dt}=cy$ 

Here $\text{c}$ is constant.

Simplify the expression:
$\frac{dy}{y}=cdt$ 

Integrate both side:
$\int \frac{dy}{y}=\int cdt$ 

$\Rightarrow \log y=ct+D$ 

$\Rightarrow y={{e}^{ct+D}}$ 

$\Rightarrow y=A{{e}^{ct}}\,\,\left( A={{e}^{D}} \right)$ 

At $\text{t=0}$, $\text{y=100000}$: 

$100000=A{{e}^{c\left( 0 \right)}}$ 

$\Rightarrow A=100000$ 

At $\text{t=2}$, $\text{y=}\frac{11}{10}\left( \text{100000} \right)=110000$: 

$y=100000{{e}^{ct}}$ 

$\Rightarrow 110000=100000{{e}^{c\left( 2 \right)}}$ 

$\Rightarrow {{e}^{2c}}=\frac{11}{10}$ 

$\Rightarrow 2c=\log \left( \frac{11}{10} \right)$ 

$\Rightarrow c=\frac{1}{2}\log \left( \frac{11}{10} \right)$ ……(1)

For $\text{y=200000}$:

$200000=100000{{e}^{ct}}$ 

$\Rightarrow {{e}^{ct}}=2$ 

$\Rightarrow ct=\log 2$ 

$\Rightarrow t=\frac{\log 2}{c}$ 

Back substituting using expression (1):
\[t=\frac{\log 2}{\frac{1}{2}\log \left( \frac{11}{10} \right)}\] 

$\Rightarrow t=\frac{2\log 2}{\log \left( \frac{11}{10} \right)}$ 

Thus time required for bacteria to reach $\text{200000}$ is \[\text{t=}\frac{\text{log2}}{\frac{\text{1}}{\text{2}}\text{log}\left( \frac{\text{11}}{\text{10}} \right)}\] hrs.

Ans: The given differential equation is $\frac{\text{dy}}{\text{dx}}\text{=}{{\text{e}}^{\text{x+y}}}$. Simplify the expression:

$\frac{dy}{dx}={{e}^{x}}{{e}^{y}}$ 

$\Rightarrow \frac{dy}{{{e}^{y}}}={{e}^{x}}dx$ 

$\Rightarrow {{e}^{-y}}dy={{e}^{x}}dx$ 

Integrate both side:

$\int {{e}^{-y}}dy=\int {{e}^{x}}dx$ 

$\Rightarrow -{{e}^{-y}}={{e}^{x}}+D$ 

$\Rightarrow {{e}^{x}}+{{e}^{-y}}=-D$ 

$\Rightarrow {{e}^{x}}+{{e}^{-y}}=C\,\,\left( C=-D \right)$ 

Thus the general solution of given differential equation is $ {{e}^{x}}+{{e}^{-y}}=C$ 

Thus the correct option is (A).