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Author Topic: [4] Exact Equations  (Read 33037 times)
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sammy
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« Reply #30 on: November 27, 2006, 12:59:53 AM »

problem 9

\begin{array}{rcl}(x-y^3+y^2\sin x)dx&=&(3xy^2+2y\cos x)dy\\\\(x-y^3+y^2\sin x)dx-(3xy^2+2y\cos x)dy&=&0\end{array}

solution
\begin{array}{rcl}M(x,y)&=&x-y^3+y^2\sin x\\\\N(x,y)&=&-3xy^2-2y\cos x\\\\\frac{\partial M}{\partial y}{ }={ }-3y^2+2y\sin x{ }={ }\frac{\partial N}{\partial x}\end{array}

so this is exact equation

\begin{array}{rcl}\frac{\partial f}{\partial x}&=&M(x,y)\\\\f(x,y)&=&\int M(x,y)dx+g(y)\\\\&=&\int(x-y^3+y^2\sin x)dx+g(y)\\\\&=&\frac{x^2}{2}-y^3x-y^2\cos x+g(y)\end{array}        ...........................(1)

differentiate f(x,y) with respect to y can get N(x,y)

\begin{array}{rcl}\frac{\partial}{\partial y}(\frac{x^2}{2}-y^3x-y^2\cos x+g(y)&=&-3xy^2-2y\cos x\\\\-3xy^2-2y\cos x+g^\prime(y)&=&-3xy^2-2y\cos x\\\\g^\prime(y)&=&0\end{array}

sog(y){ }={ }\text{constant}\end{array}

from (1) explicit solution is
xy^3+y^2\cos x-\frac{x^2}{2}=c

By Woradee Kongthong 4805190
« Last Edit: November 27, 2006, 08:29:52 AM by ปิยพงษ์ - Head Admin » Logged
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