Question

Solve the initial-value problem y'-(3/x)y=(2x^4(4x^3-3y))/(3x^5+3x^3+2y), y(1)=1.

Answer 1

\[y'=\frac{ 8x^7-6x^4y }{ 3x^5+3x^3+2y }+\frac{ 3 }{ x }y=\frac{ 8x^8-6x^5y+3y(3x^5+3x^3+2y) }{ x(3x^5+3x^3+2y) }\]

Answer 2

\[=\frac{ 8x^8+3x^5y+9x^3y+6y^2 }{ 3x^6+3x^4+2xy }\]

Answer 3

\[8x^8+3x^5y+9x^3y+6y^2-(3x^6+3x^4+2xy)y'=0\]

Answer 4

Check for exactness:
\[\frac{ d }{ dy }(8x^8+3x^5y+9x^3y+6y^2)=3x^5+9x^3+12y\]

Answer 5

\[\frac{\partial }{\partial x}[-(3x^6+3x^4+2xy)]=-18x^5-12x^3-2y \\ \frac{\partial }{\partial y}[8x^8+3x^5y+9x^3y+6y^2]=3x^5+9x^3+12y\]

\[\mu=\exp(\int\limits_{}^{ } \frac{(3x^5+9x^3+12y)+(18x^5+12x^3+2y)}{-(3x^6+3x^4+2xy)} dx) \\ =\exp(\int\limits_{}^{} \frac{21x^5+21x^3+14y}{-(3x^6+3x^4+2xy)}dx) \\ =\exp(\frac{7}{-1}\int\limits_{}^{} \frac{3x^5+3x^3+2y}{3x^6+3x^4+2xy}dx)\]

Answer 6

\[=\exp(-7 \int\limits_{}^{} \frac{3x^5+3x^3+2y}{x(3x^5+3x^3+2y)} dx\]

Answer 7

\[\mu=\exp(-7 \int\limits_{}^{}\frac{1}{x} dx)=\exp(-7 \ln|x|)\]

Answer 8

\[\mu=e^{\ln(x^{-7})}=x^{-7}\]

Answer 9

\[8x+3x^{-2}y+9x^{-4}y+6x^{-7}y^2-(3x^{-1}+3x^{-3}+2x^{-6}y)y'=0 \\ \frac{\partial }{ \partial x}(-(3x^{-1}+3x^{-3}+2x^{-6}y))=3x^{-2}+9x^{-4}+12x^{-7}y \\ \frac{\partial }{\partial y}(8x+3x^{-2}y+9x^{-4}y+6x^{-7}y^2)=3x^{-2}+9x^{-4}+12x^{-7}y\]

Answer 10

So we now have an exact differentiable equation

Answer 11

Thank you.