Question

КОШИ. Пожалуйста, помогите найти решение задачи Коши.

Answer 1

$x^2(2y-1)=(x^3+1)y'\\ \int\dfrac{x^2dx}{x^3+1}=\int\dfrac{dy}{2y-1}\\ \dfrac{1}{3}\int\dfrac{d(x^3+1)}{x^3+1}=\dfrac{1}{2}\int\dfrac{d(2y-1)}{2y-1}\\ \dfrac{1}{3}ln(C(x^3+1))=\dfrac{1}{2} ln(2y-1)\\ y(0)=1=>\dfrac{1}{3}ln(C)=\dfrac{1}{2} ln1=>C=1=>\\ =>\sqrt[3]{x^3+1}=\sqrt{2y-1}$

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$(x^2+2y^2)-x^2y'=0\\ \left[y=zx=>y'=z'x+z\right]\\ (x^2+2z^2x^2)-x^2(z'x+z)=0\\ 2z^2-z+1=z'x\\ \int\dfrac{dz}{2z^2-z+1}=\int\dfrac{dx}{x}\\ (*) \int\dfrac{dz}{2z^2-z+1}= \int\dfrac{dz}{(z\sqrt{2}-\dfrac{1}{2\sqrt2})^2+\dfrac{7}{8}}=\left[u=\dfrac{4z-1}{\sqrt7}=>dz=\dfrac{\sqrt{7}}{4}du\right]=$

$=\dfrac{2}{\sqrt7}\int\dfrac{du}{u^2+1}=\dfrac{2}{\sqrt7}arctgu+C= \dfrac{2}{\sqrt7}arctg\dfrac{4z-1}{\sqrt7}+C\\ \dfrac{2}{\sqrt7}arctg\dfrac{4z-1}{\sqrt7}+C=lnx\\ \dfrac{2}{\sqrt7}arctg\dfrac{4y-x}{x\sqrt7}+C=lnx\\ y(1)=\dfrac{\sqrt7}{4}=>\dfrac{2}{\sqrt7}arctg\dfrac{4*\dfrac{\sqrt7}{4}-1}{\sqrt7}+C=ln1=>\dfrac{2}{\sqrt7}arctg\dfrac{\sqrt7-1}{\sqrt7}+C=0=>C=-\dfrac{2}{\sqrt7}arctg\dfrac{\sqrt7-1}{\sqrt7}\\ \dfrac{2}{\sqrt7}arctg\dfrac{4y-x}{x\sqrt7}-\dfrac{2}{\sqrt7}arctg\dfrac{\sqrt7-1}{\sqrt7}=lnx$