Уравнение нерешено относно производната

[Гост]

Моля за помощ с решаването на следното диференциално уравнение

[tex]xy^{4}y′ + 3y^{5} + (y′)^{4} = 0.[/tex]

[ammornil]

Моля за помощ с решаването на следното диференциално уравнение

[tex]xy^{4}y′ + 3y^{5} + (y′)^{4} = 0.[/tex]

$\\[24pt] xy^{4}\dfrac{dy}{dx}+3y^{5}+\left(\dfrac{dy}{dx}\right)^{4}=0 \\[12pt] \because y=c\cdot{x^{k}}, \quad k\in{\mathbb{R}}, c\in{\mathbb{R}} \Rightarrow \quad \dfrac{dy}{dx} = k \cdot{c} \cdot{x^{k-1}} \\[12pt] x \cdot{(c\cdot{x^{k}})^{4}} \cdot{k \cdot{c} \cdot{x^{k-1}}} + 3 \cdot{(c\cdot{x^{k}})^{5}} + (k \cdot{c} \cdot{x^{k-1}})^{4} = 0 \\[6pt] k \cdot{c^{5}} \cdot{x^{5k}} +3 \cdot{c^{5}} \cdot{x^{5k}} + k^{4} \cdot{c^{4}} \cdot{x^{4k-4}} = 0 \\[6pt] (k+3) \cdot{c^{5}} \cdot{x^{5k}} + k^{4} \cdot{c^{4}} \cdot{x^{4k-4}} = 0 \\[12pt] \boxed{(1)}\quad c=0 \Rightarrow y=0 \\[12pt] \boxed{(2)} \quad c\ne{0} \Rightarrow (k+3) \cdot{c^{5}} \cdot{x^{5k}} + k^{4} \cdot{c^{4}} \cdot{x^{4k-4}} = 0 \quad |\div{c^{4}\ne{0}} \\[6pt] (k+3) \cdot{c} \cdot{x^{5k}} + k^{4} \cdot{x^{4k-4}} = 0 \Rightarrow \therefore 5k = 4k-4 \Rightarrow k=-4 \\[6pt] -c \cdot{x^{-20}} +256 \cdot{x^{-20}} = 0 \quad \Leftrightarrow \quad (256-c) \cdot{x^{-20}}=0 \quad \Rightarrow \quad c=256 \quad \Rightarrow \quad y=256 \cdot{x^{-4}} \\[24pt]$ $$ y=0 \quad \cup \quad y= \dfrac{256}{x^{4}} \\[24pt]$$

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Проверка:$\\[6pt] \boxed{(1)} \quad y=0 \Rightarrow 0 + 0 + 0 = 0 \Rightarrow 0 = 0 \\[12pt] \boxed{(2)} \quad y= \dfrac{2^{8}}{x^{4}}, x\ne{0} \quad \Rightarrow y^{4}= \dfrac{2^{32}}{x^{16}}, \quad y^{5}= \dfrac{2^{40}}{x^{20}} \\[6pt] \quad \Rightarrow y'=2^{8} \cdot{\dfrac{-4x^{3}}{x^{8}}}= -\dfrac{2^{10}}{x^{5}}, \quad (y')^{4}=\dfrac{2^{40}}{x^{20}} \\[6pt] xy^{4}y′ + 3y^{5} + (y′)^{4} = 0 \quad \Leftrightarrow \quad x \cdot{\dfrac{2^{32}}{x^{16}}} \cdot{\left( -\dfrac{2^{10}}{x^{5}} \right)} +3 \cdot{\dfrac{2^{40}}{x^{20}}} +\dfrac{2^{40}}{x^{20}} = 0 \\[6pt] \quad \Leftrightarrow \quad -\dfrac{2^{42}}{x^{20}} +3 \dfrac{2^{40}}{x^{20}} +\dfrac{2^{40}}{x^{20}} =0 \quad \Leftrightarrow \quad (-2^{2} +3 +1) \dfrac{2^{40}}{x^{20}} = 0 \Rightarrow 0 = 0$