Question

Найти общее решение однородного
дифференциального уравнения.

Answer 1

$xy'=\frac{3y^3+4yx^2}{2y^2+2x^2}\\y=tx;y'=t'x+t\\x(t'x+t)=\frac{3t^3x^3+4tx^3}{2t^2x^2+2x^2}\\t'x+t=\frac{3t^3+4t}{2t^2+2}|*2t^2+2\\(t'x+t)(2t^2+2)=3t^3+4t\\t'x(2t^2+2)+2t^3+2t=3t^3+4t\\t'x(2t^2+2)=t^3+2t|*\frac{dx}{2x(t^3+2t)}\\\frac{t^2+1}{t^3+2t}dt=\frac{dx}{x}\\t^3+2t=0\\t^2+2=0;t=0\\y=^+_-\sqrt2xi;y=0\\y'=^+_-\sqrt2i;y'=0\\\\\sqrt2xi=\frac{-2\sqrt2x^3i}{-2x^2}\\\sqrt2xi=\sqrt2xi\\\\-\sqrt2xi=\frac{2\sqrt2x^3i}{-2x^2}\\-\sqrt2xi=-\sqrt2xi\\\\0=\frac{0}{2x^2}\\0=0$
$\it\displaystyle \frac{t^2+1}{t^3+2t}dt=\frac{dx}{x}\\\frac{t^2+1}{t^3+2t}=\frac{A}{t}+\frac{Bt+C}{t^2+2}=\frac{1}{2t}+\frac{t}{2t^2+4}\\t^2+1=A(t^2+2)+(Bt+C)t\\t^2|1=A+B=\ \textgreater \ B=\frac{1}{2}\\t|0=C\\t^0|1=2A=\ \textgreater \ A=\frac{1}{2}\\\int(\frac{1}{2t}+\frac{t}{2(t^2+2)})dt=\int\frac{dx}{x}\\\frac{1}{2}ln|t|+\frac{1}{4}ln|t^2+2|=ln|x|+C\\\frac{1}{2}ln|\frac{y}{x}|+\frac{1}{4}ln|\frac{y^2+2x^2}{x^2}|-ln|x|=C;\\y=^+_-\sqrt2ix;\\y=0$
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$\it\displaystyle y'=\frac{x+2y}{2x-y}\\y=tx;y'=t'x+t\\t'x+t=\frac{x+2tx}{2x-tx}\\t'x+t=\frac{1+2t}{2-t}\\t'x=\frac{1+2t-2t+t^2}{2-t}\\t'x=\frac{t^2+1}{2-t}|*\frac{dx(2-t)}{x(t^2+1)}\\\frac{2-t}{t^2+1}dt=\frac{dx}{x}\\t^2+1=0\\t^2=-1\\t=^+_-i\\y=^+_-ix\\y'=^+_-i\\\\i=\frac{1+2i}{2-i}\\1+2i=1+2i\\\\-i=\frac{1-2i}{2+i}\\1-2i=1-2i\\\\\\\int(\frac{2}{t^2+1}-\frac{t}{t^2+1})dt=\int\frac{dx}{x}\\2arctgt-\frac{1}{2}ln|t^2+1|=ln|x|+C\\2arctg\frac{y}{x}-\frac{1}{2}ln|\frac{y^2+x^2}{x^2}|-ln|x|=C;y=^+_-ix$
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$\it\displaystyle xy'=2\sqrt{x^2+y^2}+y\\y=tx;y'=t'x+t\\x(t'x+t)=2\sqrt{x^2+x^2t^2}+tx|:x\\t'x+t=2\sqrt{1+t^2}+t\\t'x=2\sqrt{1+t^2}|*\frac{dx}{x\sqrt{1+t^2}}\\\frac{dt}{\sqrt{1+t^2}}=2\frac{dx}{x}\\\\\\1+t^2=0\\t^2=-1\\t=^+_-i\\y=^+_-ix\\y'=^+_-i\\\\ix=2\sqrt{x^2-x^2}+ix\\ix=ix\\\\-ix=2\sqrt{x^2-x^2}-ix\\-ix=-ix\\\\\int\frac{dt}{\sqrt{1+t^2}}=2\int\frac{dx}{x}\\ln|t+\sqrt{1+t^2}|=2ln|x|+C\\ln|\frac{y}{x}+\sqrt{\frac{y^2+x^2}{x^2}}|-2ln|x|=C\\ln|\frac{y+\sqrt{y^2+x^2}}{x}|-2ln|x|=C$
$\it\displaystyle ln|y+\sqrt{y^2+x^2}|-3ln|x|=C;y=^+_-ix$