Consider the fifferential equation dy/dx=(3-y)cosx.Let y=f(x)be the particular solution to the differential equation with the inital condition f(0)=1.

Answer 1

dy / dx = ( 3 - y ) cos x

Divide both sides by 3 - y

dy / [ dx ∙ ( 3 - y ) ] = cos x

Multiply both sides by dx

dy / ( 3 - y ) = cos x dx

Substitution:

3 - y = u

- dy = du

dy = - du

∫ dy / ( 3 - y ) = ∫ - du / u = - ∫ du / u = - log ( u ) + C₁ = - log ( 3 - y ) + C₁

∫ cos x dx = sin x + C₂

∫ dy / ( 3 - y ) = ∫ cos x dx

- log ( 3 - y ) + C₁ = sin x + C₂

Subtract C₁ to both sides

- log ( 3 - y ) = sin x + C₂ - C₁

C₂ - C₁ = C

- log ( 3 - y ) = sin x + C

Multiply both sides by - 1

log ( 3 - y ) = - sin x - C

e^[ log ( 3 - y ) ] = e^( - sin x - C )

3 - y = e^(- sin x ) ∙ e^( - C )

Subtract 3 to both sides

- y = e^( - sin x ) ∙ e^( - C ) - 3

Multiply both sides by - 1

y = - e^( - sin x ) ∙ e^( - C ) + 3

x = 0 , y = 1

y = - e^( - sin x ) ∙ e^( - C ) + 3

1 = - e^( - sin 0 ) ∙ e^( - C ) + 3
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- sin 0 = - ( 0 ) = 0
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1 = - e^0 ∙ e^( - C ) + 3

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e^0 = 1
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1 = - 1 ∙ e^( - C ) + 3

1 = - e^( - C ) + 3

Add e^( - C ) to both sides

1 + e^( - C ) = - e^( - C ) + e^( - C ) + 3

1 + e^( - C ) = 3

Subtract 1 to both sides

e^( - C ) = 2

y = - e^( - sin x ) ∙ e^( - C ) + 3

y = - e^( - sin x ) ∙ 2 + 3

y = - 2 e^( - sin x ) + 3