Question

dy/dx+(y-1)cosx=e to the power -sinx cos²x​

Answer 1

Question :-

$\sf \red{ \frac{dy}{dx} + (y - 1) \cos x} = \green{{e}^{ - \sin x} { \cos}^{2} xdx}$

Answer :-

$\boxed{ \red{ \sf y {e}^{ \sin x} } = \frac{x}{2} + \green{ \frac{ \sin 2x}{4} + {e}^{ \sin x} } + c} \\ \:$

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) .

$\to \: \sf \large \green{ \: i.f = {e}^{ \int p \: dx} }$

and after that ,

we will get our solution by -

$\to \sf \large \: y \:( i.f) = \blue{ \int \: (i.f) \: q \: dx}$

Explanation :-

firstly separate it like given equation,

$\to \sf \large \: \frac{dy}{dx} + y \cos x = {e}^{ - \sin x} \: { \cos}^{2} x + \cos x$

Now

Find integrating factor ,

$\to \sf \large i.f = {e}^{ \int \cos x \: dx} \\ \\ \to \large \sf \: \: i.f = {e}^{ \sin x}$

NOW ,

we will got our solution by -

$\to \sf \red{ y {e}^{ \sin x} } = \blue{ \int {e}^{ \sin x}} \green{ ( {e}^{ - \sin x} { \cos}^{2} x + } \cos x)dx \\ \\ \sf \red{ \to y \: {e}^{ \sin x}} = \int { \cos}^{2} x \blue{ dx + \int {e}^{ \sin x} \cos x \: } dx \\ \\ \to \sf \orange{ y {e}^{ \sin x} = \frac{1}{2} } \red{ \int(1 + \cos2x)dx + \int {e}^{t} dt} \\ \\ \to \sf y {e}^{ \sin x} = \green{\frac{x}{2} + \frac{ \sin 2x}{4}} + {e}^{t} + c \\ \\ \to \boxed{ \red{ \sf y {e}^{ \sin x} } = \frac{x}{2} + \green{ \frac{ \sin 2x}{4} + {e}^{ \sin x} } + c}$

Hope it will help you.

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Answer 2

Answer:

Step-by-step explanation: