Differential Equations. MEDAL!!!
if the slope of a curve y=f(x) is \[\frac{ x+y }{ -x }\] at all points. given the point (1,3) find the equation of the curve
\[f'(x)=y'=-\dfrac{x+y}{x}\] \[-y'x=-\frac{dy}{dx}x=x+y\implies -\frac{dy}{dx}=1+\dfrac{y}{x}\] \[tx=y\] \[-(\dfrac{dt}{dx}x+t)=1+t\] \[\dfrac{dt}{dx}x=-2t\implies\dfrac{dt}{2t}=-\dfrac{dx}{x}\] \[\ln 2t=\ln 2(\dfrac{y}{x})=\ln x+c\] plug in (1,3) to get c
thanks a lot
\[\frac{ dy }{dx }=-\frac{ x+y }{x}=-1-\frac{ y }{x }\] \[\frac{ dy }{dx }+\frac{ 1 }{ x }y=-1\] \[I.F=e ^{\int\limits \frac{ 1 }{ x }dx}=e ^{\ln x}=x\] c.s. is \[y*x=\int\limits -x~dx=\frac{ -x^2 }{2 }+c\] when x=1,y=3 \[3*1=-\frac{ 1^2 }{ 2 }+c\] \[or~c=3+\frac{ 1 }{2 }=\frac{ 7 }{2 }\] hence eq. of curve is \[yx=\frac{ -x^2 }{2 }+\frac{ 7 }{2 }\] \[or~2xy=7-x^2\]
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