Question

Решить:
(1 + х) у dх = (у – 1) dу, если у = 1 при х = 1

Answer 1

$(1+x)y\, dx=(y-1)\, dy\\\\\int (1+x)\, dx=\int \frac{(y-1)\, dy}{y}\\\\\int (1+x)\, dx=\int (1-\frac{1}{y})\, dy\\\\x+\frac{x^2}{2}=y-ln|y|+C\\\\y(1)=1:\; \; 1+\frac{1}{2}=1-ln1+C\; ,\; \; C=\frac{1}{2}\; \; \Rightarrow \\\\x+\frac{x^2}{2}=y-ln|y|+\frac{1}{2}$