Question

solve please in detail and forward me please​

Answer 1

Step-by-step explanation:

Answer is in the pic.

hope it will help you !

Answer 2

❏ QuesTion :-

Solve this differential equation

$\implies \sf \frac{dy}{dx} + \frac{y}{x log(x) } = \frac{ \sin2x}{ log(x) }$

❏ Answer :-

$\boxed{\to \orange{ \sf y \: log(x) = } \sf \frac{ -\cos2x}{2} + c} \:$

✦SoluTion :-

• Used Concept :-

We have a linear differential equation .

Process to solve -

we have a standard linear equation

$\to \sf \large \orange{ \frac{dy}{dx} } + \red{p \: y = q}$

Here P and q are the functions of "x",

Here we defined a term called " Integrating factor (i.f) .

$\to \: \sf \large \green{ \: i.f = {e}^{ \int p \: dx} } \\ \\ \bf After \: this \\ \\ \sf \red{y (i.f) = \int (i.f)\: q \: dx }$

✦ExplanaTion :-

According to the given concept we have

$\to \sf \: p = \frac{1}{x log(x) } \\ \\ \to \sf \: integrating factor(i.f) = {e}^{ \int p \: dx} \\ \\ \to \sf \: i.f = {e}^{ \int \frac{1}{x log(x) } \: dx } \\ \\ \to \sf put \: \log x \: = t \\ \\ \sf differentiate \: with \: respect \: to \: x \\ \\ \to \sf \: \frac{1}{x} \: dx = dt \\ \\ \red{ \bf \: now \: we \: have \: } \\ \\ \to \sf i.f = {e}^{ \int \: \frac{1}{t} \: dt } \\ \\ \to \sf \: i.f = {e}^{ log( log(x) ) } \\ \\ \sf \to \boxed{ \sf i.f = log(x) \: }$

Now according to the given concept -

$\to \sf \: \red{y \: (i.f) }= \green{ \int \: (i.f) \: q \: dx} \\ \\ \to \sf y \: log(x) = \purple{ \int log(x) \times \frac{ \sin2x}{ log(x) } \: dx} \\ \\ \to \red{ \sf y \: log(x) } = \blue{ \int \sin2x \: dx} \\ \\ \boxed{\to \orange{ \sf y \: log(x) = } \sf \frac{ -\cos2x}{2} + c}$

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Hope it helps you .