Answer to Question #138858 in Differential Equations for Shamsa Rahman

Question

dx=(10y+6z)dy+6y, dx=(3y2+4yz)dy+(2yz+y2)dz, dx=y4-1dy+zdz

Answer 1

given,

$dx=(10y+6z)dy+6y$

Integrate both side we get,

$x=5y^2+6yz+6y+c$,

dx=$(3y^2+4yz)dy+(2yz+y^2)dz$

This is of the form

Mdy+Ndz

so the solution is,

Integrate both the sides,

$x=y^3+2zy^2+c$ ,

dx=$(y^4-1)dy+zdz$

Integrate both the sides,

$x=\frac{y^5}{5}-y+\frac{z^2}{2}+c$