Question

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Answer 1

GIVEN :–

$\\ \to \bf \: x \dfrac{dy}{dx} + y = x \cos(x) + \sin(x)$

• And –

$\\ \to \bf \: y = 1 \: \: when \: \: x = \dfrac{\pi}{2}$

TO FIND :–

• Solution of Differential equation = ?

SOLUTION :–

$\\ \implies\bf \: x \dfrac{dy}{dx} + y = x \cos(x) + \sin(x)$

$\\ \implies\bf \: \dfrac{dy}{dx} + \dfrac{y}{x} = \cos(x) + \dfrac{ \sin(x)}{x}$

• Standard equation –

$\\ \implies\bf \: \dfrac{dy}{dx} +Py =Q$

• Now let's compare –

$\\ \implies\bf P = \dfrac{1}{x},Q =\cos(x)+ \dfrac{ \sin(x)}{x}$

• Calculating I.F. (Integrating factor) :–

$\\ \implies\bf I.F. = {e}^{ \int P.dx}$

$\\ \implies\bf I.F. = {e}^{\int \frac{1}{x} .dx}$

$\\ \implies\bf I.F. = {e}^{ log(x) }$

$\\ \implies\large{ \boxed{ \bf I.F.=x }}$

Genral solution :–

$\\ \implies\bf y(I.F.) = \int Q(I.F.).dx \: + c$

$\\ \implies\bf y(x) = \int \left(\cos(x)+ \dfrac{ \sin(x)}{x} \right)(x).dx \: + c$

$\\ \implies\bf y(x) = \int \{x\cos(x)+ \sin(x) \}.dx+ c$

$\\ \implies\bf y(x) = \int d(x \sin x)+ c$

$\\ \implies\bf y(x) = x \sin x+ c$

• According to the question –

$\\ \to \bf \: y = 1 \: \: when \: \: x = \dfrac{\pi}{2}$

• So that –

$\\ \implies\bf (1). \dfrac{\pi}{2} = \dfrac{\pi}{2} \sin \left( \dfrac{\pi}{2} \right)+ c$

$\\ \implies\bf \dfrac{\pi}{2} = \dfrac{\pi}{2}+ c$

$\\ \implies \large{ \boxed{\bf c = 0}}$

• Hence the equation is –

$\\ \implies\bf y(x) = x \sin x$

$\\ \implies \large{ \boxed{\bf y= \sin(x)}}$

Answer 2

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