Question

giải phương trình (1+y^2)(e^2xdx-e^ydy)-(1+y)dy=0

Answer 1

Step-by-step explanation:

We have,

$(1 + {y}^{2} )( {e}^{2x} dx - {e}^{y} dy) - (1 + y)dy = 0$

$\implies {e}^{2x} dx - {e}^{y} dy - \bigg(\frac{1 + y}{1 + {y}^{2} } \bigg)dy = 0$

$\implies \int {e}^{2x} dx - \int{e}^{y} dy - \int \bigg(\frac{1 + y}{1 + {y}^{2} } \bigg)dy = 0$

$\implies \frac{ {e}^{2x}}{2} - {e}^{y} - \int \frac{1}{1 + {y}^{2} }dy - \int \frac{y}{1 + {y}^{2} }dy = 0$

$\implies \frac{ {e}^{2x}}{2} - {e}^{y} - \tan^{ - 1} (y) - \int \frac{y}{1 + {y}^{2} }dy = 0$

$\implies \frac{ {e}^{2x}}{2} - {e}^{y} - \tan^{ - 1} (y) - \frac{1}{2} \int \frac{2y}{1 + {y}^{2} }dy = 0$

$\implies \frac{ {e}^{2x}}{2} - {e}^{y} - \tan^{ - 1} (y) - \frac{1}{2} ln | 1 + {y}^{2}| = 0$

$\implies \frac{ {e}^{2x}}{2} - {e}^{y} - \tan^{ - 1} (y) - ln \sqrt{1 + {y}^{2}}= 0$