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Question:

$y^{\prime}=\frac{x+y}{x}$

Solution:

The given differential equation is:

$y^{\prime}=\frac{x+y}{x}$

$\Rightarrow \frac{d y}{d x}=\frac{x+y}{x}$            …(1)

Let $F(x, y)=\frac{x+y}{x}$.

Now, $F(\lambda x, \lambda y)=\frac{\lambda x+\lambda y}{\lambda x}=\frac{x+y}{x}=\lambda^{0} F(x, y)$

Thus, the given equation is a homogeneous equation.

To solve it, we make the substitution as:

vx

Differentiating both sides with respect to x, we get:

$\frac{d y}{d x}=v+x \frac{d v}{d x}$

Substituting the values of $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=\frac{x+v x}{x}$

$\Rightarrow v+x \frac{d v}{d x}=1+v$

$x \frac{d v}{d x}=1$

$\Rightarrow d v=\frac{d x}{x}$

Integrating both sides, we get:

$v=\log x+\mathrm{C}$

$\Rightarrow \frac{y}{x}=\log x+\mathrm{C}$

$\Rightarrow y=x \log x+\mathrm{C} x$

This is the required solution of the given differential equation.

 

 

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