Author Topic: TUT0402 Quiz2  (Read 317 times)

Jingjing Cui

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TUT0402 Quiz2
« on: October 04, 2019, 02:00:03 PM »
$$
\\ (3x^2y+2xy+y^3)+(x^2+y^2)y'=0
\\ M=(3x^2y+2xy+y^3)
\\ N=(x^2+y^2)
\\ My=3x^2+2x+3y^2
\\ Nx=2x
\\ R2=\frac{My-Nx}{N}=\frac{3x^2+2x+3y^2-2x}{(x^2+y^2)}=3
\\
\\ \mu=e^{\int{R2}dx}=e^{\int{3}dx}=e^{3x}
\\
\\ Multiply\; both\; sides\; by\; \mu\; ,\; we\; get\;
\\ e^{3x}(3x^2y+2xy+y^3)+e^{3x}(x^2+y^2)\frac{dy}{dx}=0
\\ M1=e^{3x}(3x^2y+2xy+y^3)
\\ N1=e^{3x}(x^2+y^2)
\\
\\ \phi=\int{N1}dy=\int{e^{3x}(x^2+y^2) }dy=e^{3x}x^2y+\frac{1}{3}e^{3x}y^3+h(x)
\\ \phi{x}=3e^{3x}x^2y+2e^{3x}xy+e^{3x}y^3+h'(x)=e^{3x}(3x^2y+2xy+y^3)+h'(x)=M1
\\ Therefore\; ,\; h'(x)=0\; h(x)=C
\\ \phi=e^{3x}x^2y+\frac{1}{3}e^{3x}y^3+C
\\ Thus\; ,\; the\; solution\; is\; C=e^{3x}x^2y+\frac{1}{3}e^{3x}y^3
$$