Question
Solve the following initial value problems:
$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=\log\text{x},\text{ y}(1)=0$
$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=\log\text{x},\text{ y}(1)=0$
✓
Answer
We have,
$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=\log\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}-\frac{\text{y}}{\text{x}}=\frac{\log\text{x}}{\text{x}}\ ...(\text{1})$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
Where $\text{P}=-\frac{1}{\text{x}}$ and $\text{Q}=\frac{\log\text{x}}{\text{x}}$
$\therefore\text{I.F.}=\text{e}^{\int\text{Pdx}}$
$=\text{e}{-\int\frac{1}{\text{x}}}\text{ dx}$
$=\text{e}^{-\log\text{x}}$
$=\frac{1}{\text{x}}$
Multiplying both sides of (1) by $\text{I.F.}=\frac{1}{\text{x}},$ we get
$\frac{1}{\text{x}}\Big(\frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}}\text{y}\Big)=\frac{1}{\text{x}}\times\frac{\log\text{x}}{\text{x}}$
$\Rightarrow\frac{1}{\text{x}}\frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}^2}\text{y}=\frac{\log\text{x}}{\text{x}^2}$
Integrsting both sides with respect to x, we get
$\text{y}\frac{1}{\text{x}}=\int\frac{1}{\text{x}^2}\times\log\text{x dx}+\text{C}$
$\Rightarrow\frac{\text{y}}{\text{x}}=\log\text{x}\int\frac{1}{\text{x}^2}\text{dx}-\int\Big[\frac{\text{d}}{\text{dx}}(\log\text{x})\int\frac{1}{\text{x}^2}\text{dx}\Big]\text{dx}+\text{C}$
$\Rightarrow\frac{\text{y}}{\text{x}}=-\frac{\log\text{x}}{\text{x}}+\int\frac{1}{\text{x}^2}\text{dx}+\text{C}$
$\Rightarrow\frac{\text{y}}{\text{x}}=-\frac{\log\text{x}}{\text{x}}-\frac{1}{\text{x}}+\text{C}$
$\Rightarrow\text{y}=-\log\text{x}-1+\text{Cx}\ ...(\text{ii})$
Now,
$\text{y}(1)=0$
$\therefore\ 0=-0-1+\text{C}(1)$
$\Rightarrow\text{C}=1$
Putting the value of C in (2) we get
$\text{y}=-\log\text{x}-1+\text{x}$
$\Rightarrow\text{y}=\text{x}-1-\log\text{x}$
Hence, $\text{y}=\text{x}-1-\log\text{x}$ is the required solution.
$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=\log\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}-\frac{\text{y}}{\text{x}}=\frac{\log\text{x}}{\text{x}}\ ...(\text{1})$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
Where $\text{P}=-\frac{1}{\text{x}}$ and $\text{Q}=\frac{\log\text{x}}{\text{x}}$
$\therefore\text{I.F.}=\text{e}^{\int\text{Pdx}}$
$=\text{e}{-\int\frac{1}{\text{x}}}\text{ dx}$
$=\text{e}^{-\log\text{x}}$
$=\frac{1}{\text{x}}$
Multiplying both sides of (1) by $\text{I.F.}=\frac{1}{\text{x}},$ we get
$\frac{1}{\text{x}}\Big(\frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}}\text{y}\Big)=\frac{1}{\text{x}}\times\frac{\log\text{x}}{\text{x}}$
$\Rightarrow\frac{1}{\text{x}}\frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}^2}\text{y}=\frac{\log\text{x}}{\text{x}^2}$
Integrsting both sides with respect to x, we get
$\text{y}\frac{1}{\text{x}}=\int\frac{1}{\text{x}^2}\times\log\text{x dx}+\text{C}$
$\Rightarrow\frac{\text{y}}{\text{x}}=\log\text{x}\int\frac{1}{\text{x}^2}\text{dx}-\int\Big[\frac{\text{d}}{\text{dx}}(\log\text{x})\int\frac{1}{\text{x}^2}\text{dx}\Big]\text{dx}+\text{C}$
$\Rightarrow\frac{\text{y}}{\text{x}}=-\frac{\log\text{x}}{\text{x}}+\int\frac{1}{\text{x}^2}\text{dx}+\text{C}$
$\Rightarrow\frac{\text{y}}{\text{x}}=-\frac{\log\text{x}}{\text{x}}-\frac{1}{\text{x}}+\text{C}$
$\Rightarrow\text{y}=-\log\text{x}-1+\text{Cx}\ ...(\text{ii})$
Now,
$\text{y}(1)=0$
$\therefore\ 0=-0-1+\text{C}(1)$
$\Rightarrow\text{C}=1$
Putting the value of C in (2) we get
$\text{y}=-\log\text{x}-1+\text{x}$
$\Rightarrow\text{y}=\text{x}-1-\log\text{x}$
Hence, $\text{y}=\text{x}-1-\log\text{x}$ is the required solution.
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