Question-99286

Question Number 99286 by bobhans last updated on 20/Jun/20

Commented by bemath last updated on 20/Jun/20

$$\mathrm{Bernoulli}\:\mathrm{equation}\: \\ $$

Commented by bemath last updated on 20/Jun/20

y′ + (1/x)y = ((cos x)/y^2 ) y′+ (1/x)y = (cos x ).y^(−2) let u = y^(1−(−2)) = y^3 (du/dx) = 3y^2 (dy/dx) , (dy/dx) =(1/3) y^(−2) (du/dx) ⇔(1/3)y^(−2) (du/dx) + (1/x)y = (cos x).y^(−2) (du/dx) + (3/x)u = cos x IF v (x) = e^(∫ (3/x) dx) = e^(3 ln(x)) = x^3 ⇔u(x) = ((∫ x^3 cos x dx +C)/x^3 ) x^3 y^3 = x^3 sin x+3x^2 cos x−6xsin x−6cos x +C y^3 = sin x + ((3cos x)/x)−((6sin x)/x^2 )−((6cos x)/x^3 )+Cx^(−3)

$$\mathrm{y}'\:+\:\frac{\mathrm{1}}{\mathrm{x}}\mathrm{y}\:=\:\frac{\mathrm{cos}\:\mathrm{x}}{\mathrm{y}^{\mathrm{2}} } \\ $$$$\mathrm{y}'+\:\frac{\mathrm{1}}{\mathrm{x}}\mathrm{y}\:=\:\left(\mathrm{cos}\:\mathrm{x}\:\right).\mathrm{y}^{−\mathrm{2}} \\ $$$$\mathrm{let}\:\mathrm{u}\:=\:\mathrm{y}^{\mathrm{1}−\left(−\mathrm{2}\right)} \:=\:\mathrm{y}^{\mathrm{3}} \: \\ $$$$\frac{\mathrm{du}}{\mathrm{dx}}\:=\:\mathrm{3y}^{\mathrm{2}} \:\frac{\mathrm{dy}}{\mathrm{dx}}\:,\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{y}^{−\mathrm{2}} \:\frac{\mathrm{du}}{\mathrm{dx}} \\ $$$$\Leftrightarrow\frac{\mathrm{1}}{\mathrm{3}}\mathrm{y}^{−\mathrm{2}} \:\frac{\mathrm{du}}{\mathrm{dx}}\:+\:\frac{\mathrm{1}}{\mathrm{x}}\mathrm{y}\:=\:\left(\mathrm{cos}\:\mathrm{x}\right).\mathrm{y}^{−\mathrm{2}} \\ $$$$\frac{\mathrm{du}}{\mathrm{dx}}\:+\:\frac{\mathrm{3}}{\mathrm{x}}\mathrm{u}\:=\:\mathrm{cos}\:\mathrm{x}\: \\ $$$$\mathrm{IF}\:\mathrm{v}\:\left(\mathrm{x}\right)\:=\:\mathrm{e}^{\int\:\frac{\mathrm{3}}{\mathrm{x}}\:\mathrm{dx}} =\:\mathrm{e}^{\mathrm{3}\:\mathrm{ln}\left(\mathrm{x}\right)} \:=\:\mathrm{x}^{\mathrm{3}} \\ $$$$\Leftrightarrow\mathrm{u}\left(\mathrm{x}\right)\:=\:\frac{\int\:\mathrm{x}^{\mathrm{3}} \mathrm{cos}\:\mathrm{x}\:\mathrm{dx}\:+\mathrm{C}}{\mathrm{x}^{\mathrm{3}} } \\ $$$$\mathrm{x}^{\mathrm{3}} \mathrm{y}^{\mathrm{3}} \:=\:\mathrm{x}^{\mathrm{3}} \mathrm{sin}\:\mathrm{x}+\mathrm{3x}^{\mathrm{2}} \mathrm{cos}\:\mathrm{x}−\mathrm{6xsin}\:\mathrm{x}−\mathrm{6cos}\:\mathrm{x}\:+\mathrm{C} \\ $$$$\mathrm{y}^{\mathrm{3}} \:=\:\mathrm{sin}\:\mathrm{x}\:+\:\frac{\mathrm{3cos}\:\mathrm{x}}{\mathrm{x}}−\frac{\mathrm{6sin}\:\mathrm{x}}{\mathrm{x}^{\mathrm{2}} }−\frac{\mathrm{6cos}\:\mathrm{x}}{\mathrm{x}^{\mathrm{3}} }+\mathrm{Cx}^{−\mathrm{3}} \\ $$

Commented by bobhans last updated on 20/Jun/20

$$\mathrm{great} \\ $$

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