Name: Linear Differential Equation: Illustration
Uploaded: 2019-07-19
Duration: 3 min 43 s
Description: You might be familiar with linear equations right from your middle school. Let's upgrade the concept to linear differential equations. This video teaches you...

Question

Solve the differential equation:

x \cdot \log x \cdot \frac{d y}{d x} + y = \frac{2}{x} \cdot \log x

.

Accepted Answer

The given differential equation is $x \cdot \log x \cdot \frac{d y}{d x} + y = \frac{2}{x} \cdot \log x$

$\Rightarrow \frac{d y}{d x} + \frac{y}{x \cdot l o g x} = \frac{2}{x^{2}}$

This is the form of a linear differential equation $\frac{d y}{d x} + P y = Q$

On comparing, we get $P = \frac{1}{x \cdot \log x} and Q = \frac{2}{x^{2}}$
Now, the integrating factor, $I . F = e^{\int P d x} = e^{\int \frac{1}{x \cdot l o gx} d x} = e^{\int (\frac{\frac{1}{x}}{logx}) d x} = e^{\log (\log x)} = l o g x$ [Using $\int \frac{f^{'} (x)}{f (x)} d x = \log |f (x)| + c$ ]

The general solution of the given differential equation is given by
$(y \times I . F) = \int (Q \times I . F) d x + C$
$\Rightarrow y \times l o g x = \int (\frac{2}{x^{2}} \times l o g x) d x + C$

$\Rightarrow y \times \log x = 2 \int (l o g x \times \frac{1}{x^{2}}) d x + C$

$\Rightarrow y \times \log x = 2 [l o g x \cdot \int \frac{1}{x^{2}} d x - \int \{\frac{d}{d x} (l o g x) \cdot \int \frac{1}{x^{2}} d x\} d x] + C$ [Using integration by parts]

$\Rightarrow y \times \log x = 2 [l o g x (- \frac{1}{x}) - \int (\frac{1}{x} \cdot (- \frac{1}{x})) d x] + C$

$\Rightarrow y \times \log x = 2 [- \frac{l o g x}{x} + \int \frac{1}{x^{2}} d x] + C$

$\Rightarrow y \times \log x = 2 [- \frac{l o g x}{x} - \frac{1}{x}] + C$

$\Rightarrow y \times \log x = - \frac{2}{x} (1 + l o g x) + C$

Hence, the required general solution of the given differential equation is

$y \times l o g x = - \frac{2}{x} (1 + l o g x) + C$ .

Solve the differential equation: $x \cdot \log x \cdot \frac{d y}{d x} + y = \frac{2}{x} \cdot \log x$ .

Important Questions on Differential Equations

Learn and Prepare for any exam you want !

Solve the differential equation: x·logx·dydx+y=2x·logx.

Important Questions on Differential Equations

Learn and Prepare for any exam you want !

Solve the differential equation: $x \cdot \log x \cdot \frac{d y}{d x} + y = \frac{2}{x} \cdot \log x$ .