Question

find the integral factor dy/dx+y/x=3sinx​

Answer 1

Answer:

x

Step-by-step explanation:

Given :

$\mathrm{\dfrac{dy}{dx} + \dfrac{y}{x} = 3sinx}$

To find :

Integral factor

Solution :

We know that, if DE is of the form, $\boxed{\mathrm{\mathrm{\dfrac{dy}{dx} + P(x).y = Q(x)}}}$, then, $\boxed{\mathrm{IF = e^{\large{\mathrm{\int P(x) \, dx }}}}}$

IF = Integrating factor

$\longmapsto \mathrm{\dfrac{dy}{dx} + \dfrac{y}{x} = 3sinx}$

Here, P(x) = 1/x ; Q(x) = 3sinx

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

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

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

$\therefore \underline{\boxed{\mathrm{Integrating\ Factor = x}}}}$

❖ 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:

Step-by-step explanation:

Given :

\mathrm{\dfrac{dy}{dx} + \dfrac{y}{x} = 3sinx}

dx

dy

+

x

y

=3sinx

To find :

Integral factor

Solution :

We know that, if DE is of the form, \boxed{\mathrm{\mathrm{\dfrac{dy}{dx} + P(x).y = Q(x)}}}

dx

dy

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

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

IF=e

∫P(x)dx

IF = Integrating factor

\longmapsto \mathrm{\dfrac{dy}{dx} + \dfrac{y}{x} = 3sinx}⟼

dx

dy

+

x

y

=3sinx

Here, P(x) = 1/x ; Q(x) = 3sinx

\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

\therefore \underline{\boxed{\mathrm{Integrating\ Factor = x}}}}

❖ 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/45174871]

Hope it helps!!