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solve differential equation(x^2 + y^2 + 2x)dx + 2ydy = 0 solve differential equation(x^2 + y^2 + 2x)dx + 2ydy = 0 @Mathematics
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\[(x^2 + y^2 + 2x)dx + 2ydy = 0\]\[x^2 + y^2 + 2x + 2yy' = 0\]\[x^2 + y^2 + 2x + (y^2)'= 0\quad,\quad z=y^2\]\[z'+z=-x^2-2x\]\[z_h=Ce^{-x}\]\[z_p=ax^2+bx+c\quad,\quad z'_p=2ax+b\]\[z'_p+z_p=ax^2+(2a+b)x+b+c=-x^2-2x\quad,\quad a=-1,b=0,c=0\]\[z_p=-x^2\]\[z=y^2=Ce^{-x}-x^2\quad,\quad x^2+y^2=Ce^{-x}\]
I have this equation done and the result is \[y ^{2}=Ce ^{-x}+x ^{2}-x+1\]
what is \[z _{h}\] and \[z _{p}\]?
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