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GIVEN :–
[tex] \\ \to \bf \: x \dfrac{dy}{dx} + y = x \cos(x) + \sin(x) \\ [/tex]
• And –
[tex] \\ \to \bf \: y = 1 \: \: when \: \: x = \dfrac{\pi}{2} \\ [/tex]
TO FIND :–
• Solution of Differential equation = ?
SOLUTION :–
[tex] \\ \implies\bf \: x \dfrac{dy}{dx} + y = x \cos(x) + \sin(x) \\ [/tex]
[tex] \\ \implies\bf \: \dfrac{dy}{dx} + \dfrac{y}{x} = \cos(x) + \dfrac{ \sin(x)}{x} \\ [/tex]
• Standard equation –
[tex] \\ \implies\bf \: \dfrac{dy}{dx} +Py =Q \\ [/tex]
• Now let's compare –
[tex] \\ \implies\bf P = \dfrac{1}{x},Q =\cos(x)+ \dfrac{ \sin(x)}{x} \\ [/tex]
• Calculating I.F. (Integrating factor) :–
[tex] \\ \implies\bf I.F. = {e}^{ \int P.dx} \\ [/tex]
[tex] \\ \implies\bf I.F. = {e}^{\int \frac{1}{x} .dx} \\ [/tex]
[tex] \\ \implies\bf I.F. = {e}^{ log(x) } \\ [/tex]
[tex] \\ \implies\large{ \boxed{ \bf I.F.=x }}\\ [/tex]
Genral solution :–
[tex] \\ \implies\bf y(I.F.) = \int Q(I.F.).dx \: + c\\ [/tex]
[tex] \\ \implies\bf y(x) = \int \left(\cos(x)+ \dfrac{ \sin(x)}{x} \right)(x).dx \: + c\\ [/tex]
[tex] \\ \implies\bf y(x) = \int \{x\cos(x)+ \sin(x) \}.dx+ c\\ [/tex]
[tex] \\ \implies\bf y(x) = \int d(x \sin x)+ c\\ [/tex]
[tex] \\ \implies\bf y(x) = x \sin x+ c\\ [/tex]
• According to the question –
[tex] \\ \to \bf \: y = 1 \: \: when \: \: x = \dfrac{\pi}{2} \\ [/tex]
• So that –
[tex] \\ \implies\bf (1). \dfrac{\pi}{2} = \dfrac{\pi}{2} \sin \left( \dfrac{\pi}{2} \right)+ c\\ [/tex]
[tex] \\ \implies\bf \dfrac{\pi}{2} = \dfrac{\pi}{2}+ c\\ [/tex]
[tex] \\ \implies \large{ \boxed{\bf c = 0}}\\ [/tex]
• Hence the equation is –
[tex] \\ \implies\bf y(x) = x \sin x\\ [/tex]
[tex] \\ \implies \large{ \boxed{\bf y= \sin(x)}}\\ [/tex]
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