1 Answer

Jefflunt · Answer 1 · 2023-01-19T02:52:16+0000

Suppose we're asked to solve the first order linear differnetial equation,

$\longrightarrow\rm{(dy)/(dx)+y\cdot P(x)=Q(x)}$

To solve this equation we have to find out a term called 'Integrating Factor' given by,

$\rm{I_f=e^(\int P(x)\ dx)}$

such that the solution of the equation is,

$\displaystyle\longrightarrow\rm{y\cdot I_f=\int Q(x)\cdot I_f\ dx}$

Here the given differential equation is,

$\longrightarrow\rm{(dy)/(dx)-(y)/(x)=3x}$

$\longrightarrow\rm{(dy)/(dx)+y\left(-(1)/(x)\right)=(3x)}$

Here,

$\longrightarrow\rm{P(x)=-(1)/(x)}$

Integrating wrt x,

$\displaystyle\longrightarrow\rm{\int P(x)\ dx=-\int(1)/(x)\ dx}$

$\displaystyle\longrightarrow\rmx$

[We will consider constant of integration as zero.]

Then the integrating factor is,

$\longrightarrow\rm{I_f=e^(\int P(x)\ dx)}$

$\longrightarrow\rmx$

Since
$\rm{e^(b\,\ln(a))=a^b,}$

$\longrightarrow\rm{I_f=x^(-1)}$

$\longrightarrow\rm{\underline{\underline{I_f=(1)/(x)}}}$

Hence the integrating factor for the given differential equation is 1/x.

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