Math, asked by ankityadav12, 10 months ago

Solve this differential equation
 \frac{dy}{dx}  + (y - 1) \cos x=   {e}^{ -  \sin x}   { \cos}^{2} \times  xdx

Answers

Answered by Sharad001
72

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} \\  \:

Approach :-

ï 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} \\

Solution :-

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} \\

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