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x-dy-dx-y-y-3-cos-x-




Question Number 129334 by bramlexs22 last updated on 15/Jan/21
 x (dy/dx) = y + y^3 cos x
$$\:\mathrm{x}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{y}\:+\:\mathrm{y}^{\mathrm{3}} \mathrm{cos}\:\mathrm{x}\: \\ $$
Answered by liberty last updated on 15/Jan/21
 let y = w^(−(1/2))  →(dy/dx) = −(1/2)w^(−3/2)  (dw/dx)   ⇒−(1/2)xw^(−3/2)  (dw/dx)−w^(−1/2) =w^(−3/2) cos x   ⇒(dw/dx)+2(w/x) = −((2cos x)/x)   take integral factor ϕ=e^(∫ (dx/x)) = x   wx = ∫−((2cos x)/x).x dx = ∫−cos x dx=−2sin x+C   y^(−2)  = ((−2sin x+C)/x) or y^2 =(x/(C−2sin x))
$$\:\mathrm{let}\:\mathrm{y}\:=\:\mathrm{w}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:\rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{w}^{−\mathrm{3}/\mathrm{2}} \:\frac{\mathrm{dw}}{\mathrm{dx}} \\ $$$$\:\Rightarrow−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{xw}^{−\mathrm{3}/\mathrm{2}} \:\frac{\mathrm{dw}}{\mathrm{dx}}−\mathrm{w}^{−\mathrm{1}/\mathrm{2}} =\mathrm{w}^{−\mathrm{3}/\mathrm{2}} \mathrm{cos}\:\mathrm{x}\: \\ $$$$\Rightarrow\frac{\mathrm{dw}}{\mathrm{dx}}+\mathrm{2}\frac{\mathrm{w}}{\mathrm{x}}\:=\:−\frac{\mathrm{2cos}\:\mathrm{x}}{\mathrm{x}} \\ $$$$\:\mathrm{take}\:\mathrm{integral}\:\mathrm{factor}\:\varphi=\mathrm{e}^{\int\:\frac{\mathrm{dx}}{\mathrm{x}}} =\:\mathrm{x} \\ $$$$\:\mathrm{wx}\:=\:\int−\frac{\mathrm{2cos}\:\mathrm{x}}{\mathrm{x}}.\mathrm{x}\:\mathrm{dx}\:=\:\int−\mathrm{cos}\:\mathrm{x}\:\mathrm{dx}=−\mathrm{2sin}\:\mathrm{x}+\mathrm{C} \\ $$$$\:\mathrm{y}^{−\mathrm{2}} \:=\:\frac{−\mathrm{2sin}\:\mathrm{x}+\mathrm{C}}{\mathrm{x}}\:\mathrm{or}\:\mathrm{y}^{\mathrm{2}} =\frac{\mathrm{x}}{\mathrm{C}−\mathrm{2sin}\:\mathrm{x}} \\ $$

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