Toronto Math Forum
MAT244--2019F => MAT244--Test & Quizzes => Quiz-2 => Topic started by: huyangqu on October 04, 2019, 03:32:35 PM
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Q: (x+2)sin(y)+xcos(y)y′=0 where μ=xex
Sol:Let M(x,y)=(x+2)sin(y) , N(x,y)=xcos(y)
My=(x+2)cos(y), Nx=xcos(y)
since My and Nx is not equal, so the equation is not exact
then multiply the interal factor both sides
x²ex sin(y)+2xexsin(y) +x²excos(y)y′=0
so there ∃φ(x,y) where φx=M φy=N
so φ=∫Mdx=∫x²ex sin(y)+2xexsin(y)dx=x²ex sin(y)+h(y)
φy=x²excos(y) +h′(y)=N so h′(y)=0 therefore h(y)=c
φ=x²excos(y)+c
x²excos(y)=c is the general solution of this equation