Question

Найти интегралл с начальнім условием y(1) = 1
16xdx - 5ydy = 5x^2ydy - 4xYdx

Answer 1

$16x\, dx-5y\, dy=5x^2y\, dy-4xy\, dx\ \ \ ,\ \ \ y(1)=1\\\\(16x+4xy)\, dx=(5x^2y+5y)\, dy\\\\5y\, (x^2+1)\, dy=4x(4+y)\, dx\\\\\displaystyle \int \frac{5y\, dy}{4+y}=\int \frac{4x\, dx}{1+x^2}\\\\\\\int \Big(5-\frac{20}{y+4}\Big)\, dy=2\int \frac{2x\, dx}{1+x^2}\\\\\\5y-20\cdot ln|y+4|=2\cdot ln|1+x^2|+2\, lnC\\\\-20\cdot ln|y+4|=-5y+2\, ln(C(1+x^2))\\\\ln|y+4|=\frac{y}{4}-\frac{1}{10}\cdot ln(C(1+x^2))$

$y(1)=1:\ \ ln|1+4|=\dfrac{1}{4}-\dfrac{1}{10}\cdot ln(2C)\ \ ,\ \ \ \dfrac{1}{10}\, ln(2C)=\dfrac{1}{4}-ln5\ \ ,\\\\\\ln(2C)=\dfrac{5}{2}-ln5\ \ ,\ \ ln(2C)=2,5-ln5\ \ ,\ \ 2C=e^{2,5-ln5}=e^{2,5}\cdot \dfrac{1}{5}\ ,\\\\C=0,1\cdot e^{2,5}\\\\\\ln|y+4|=\dfrac{y}{4}-\dfrac{1}{10}\cdot ln(0,1\cdot e^{2,5}(1+x^2))\\\\0,25y-ln|y+4|=0,1\cdot ln(0,1\cdot e^{2,5}(1+x^2))$

Answer 2

Пошаговое объяснение: