Question

slove the D. E x×dy/dx+y=logx​

Answer 1

Answer:

$\mathbf{yx = xlogx - x + c }$

Step-by-step explanation:

Given Question :

$\text{To solve the given DE } \mathrm{\dfrac{xdy}{dx} + y = logx}$

Solution :

$\longmapsto \mathrm{\dfrac{xdy}{dx} + y = logx}$

Dividing by x, we get,

$\implies \mathrm{\dfrac{dy}{dx} + \dfrac{y}{x} = \dfrac{logx}{x}}$

This is of the form,

$\boxed{\mathrm{\mathrm{\dfrac{dy}{dx} + P(x).y = Q(x)}}}$

Here, P(x) = 1/x, Q(x) = logx/x

We know that, integrating factor IF is

$\boxed{\mathrm{IF = e^{\large{\mathrm{\int P(x) \, dx }}}}}$

Here, IF is

$\implies \mathrm{IF = e^{\displaystyle {\int \dfrac{1}{x}dx}}}$

$\implies \mathrm{IF = e^{logx}}$

$\implies \mathbf{\underline{IF = x}}$

We know that, solution of linear differential equation,

$\boxed{\mathrm{y.(IF) = \int Q(x).(IF) \, dx }}$

Substituting the terms, we get,

$\implies \mathrm{yx = \displaystyle \int \dfrac{logx}{x}\times x \, dx }$

$\implies \mathrm{yx = \displaystyle \int \dfrac{logx}{\not x}\times \not x \, dx }$

$\implies \mathrm{yx = \int logx\, dx }$

We know that, integration of logx is

$\boxed{\mathrm{\int logx \, dx = xlogx - x + c}}$

$\implies \mathrm{yx = xlogx - x + c }$

This is the solution of the given DE

❖ Extra information :

⟡ Some basic integrals :

$\boxed{\boxed{\begin{minipage}{4cm}\displaystyle\circ\sf\:\int{1\:dx}=x+c\\\\\circ\sf\:\int{a\:dx}=ax+c\\\\\circ\sf\:\int{x^n\:dx}=\dfrac{x^{n+1}}{n+1}+c\\\\\circ\sf\:\int{sin\:x\:dx}=-cos\:x+c\\\\\circ\sf\:\int{cos\:x\:dx}=sin\:x+c\\\\\circ\sf\:\int{sec^2x\:dx}=tan\:x+c\\\\\circ\sf\:\int{e^x\:dx}=e^x+c\end{minipage}}}$

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Answer 2

Answer:

Answer:

\mathbf{yx = xlogx - x + c }yx=xlogx−x+c

Step-by-step explanation:

Given Question :

\text{To solve the given DE } \mathrm{\dfrac{xdy}{dx} + y = logx}To solve the given DE

dx

xdy

+y=logx

Solution :

\longmapsto \mathrm{\dfrac{xdy}{dx} + y = logx}⟼

dx

xdy

+y=logx

Dividing by x, we get,

\implies \mathrm{\dfrac{dy}{dx} + \dfrac{y}{x} = \dfrac{logx}{x}}⟹

dx

dy

+

x

y

=

x

logx

This is of the form,

\boxed{\mathrm{\mathrm{\dfrac{dy}{dx} + P(x).y = Q(x)}}}

dx

dy

+P(x).y=Q(x)

Here, P(x) = 1/x, Q(x) = logx/x

We know that, integrating factor IF is

\boxed{\mathrm{IF = e^{\large{\text{$\mathrm{\int P(x) \, dx }$}}}}}

IF=e

∫P(x)dx

Here, IF is

\implies \mathrm{IF = e^{\displaystyle {\int \dfrac{1}{x}dx}}}⟹IF=e

∫

x

1

dx

\implies \mathrm{IF = e^{logx}}⟹IF=e

logx

\implies \mathbf{\underline{IF = x}}⟹

IF=x

We know that, solution of linear differential equation,

\boxed{\mathrm{y.(IF) = \int Q(x).(IF) \, dx }}

y.(IF)=∫Q(x).(IF)dx

Substituting the terms, we get,

\implies \mathrm{yx = \displaystyle \int \dfrac{logx}{x}\times x \, dx }⟹yx=∫

x

logx

×xdx

\implies \mathrm{yx = \displaystyle \int \dfrac{logx}{\not x}\times \not x \, dx }⟹yx=∫



x

logx

×



xdx

\implies \mathrm{yx = \int logx\, dx }⟹yx=∫logxdx

We know that, integration of logx is

\boxed{\mathrm{\int logx \, dx = xlogx - x + c}}

∫logxdx=xlogx−x+c

\implies \mathrm{yx = xlogx - x + c }⟹yx=xlogx−x+c

This is the solution of the given DE

❖ Extra information :

⟡ Some basic integrals :

\begin{gathered}\boxed{\boxed{\begin{minipage}{4cm}\displaystyle\circ\sf\:\int{1\:dx}=x+c\\\\\circ\sf\:\int{a\:dx}=ax+c\\\\\circ\sf\:\int{x^n\:dx}=\dfrac{x^{n+1}}{n+1}+c\\\\\circ\sf\:\int{sin\:x\:dx}=-cos\:x+c\\\\\circ\sf\:\int{cos\:x\:dx}=sin\:x+c\\\\\circ\sf\:\int{sec^2x\:dx}=tan\:x+c\\\\\circ\sf\:\int{e^x\:dx}=e^x+c\end{minipage}}}\end{gathered}

[Note : If there is any difficulty viewing this answer in app, kindly see this answer at website brainly.in through desktop mode. The link of this question is : https://brainly.in/question/45505661]

Hope it helps!!