Решите пожалуйста ду xy'=3x+4y

[latex]y'=3+4\cdot \frac{y}{x}\\\\t=\frac{y}{x},\; y=tx,\; y'=t'x+t\\\\t'x+t=3+4t\\\\t'x=3+3t\\\\t'=\frac{3(1+t)}{x}\\\\\frac{dt}{dx}=\frac{3(1+t)}{x}\\\\\int \frac{dt}{3(1+t)}=\int \frac{dx}{x}\\\\\frac{1}{3}\cdot ln|1+t|=ln|x|+lnC[/latex] [latex]\sqrt[3]{1+t}=Cx\\\\1+\frac{y}{x}=C_1x^3\\\\y=x(C_1x^3-1)[/latex]