What is 1st order homogenous differential equation?show to solve this homogenous 1st orderdifferential equation: (x^2+y^2)dx-2xy dy=0 and i want to understand the basic how can i apply homogenious 1st order differential equation term on this math because i am new with this homogenous 1st orderdifferential equation.Thanks in advance.
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What is 1st order homogenous differential equation? (x^2+y^2)dx-2xy dy=0
- Thread starter naimm
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In your class (or in your textbook) this question must have been discussed.What is 1st order homogenous differential equation?show to solve this homogenous 1st orderdifferential equation: (x^2+y^2)dx-2xy dy=0 and i want to understand the basic how can i apply homogenious 1st order differential equation term on this math because i am new with this homogenous 1st orderdifferential equation.Thanks in advance.
What did those (your classnotes and/or textbook) say?
HallsofIvy
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A first order differential equation can always be written in the form f(x,y)dy+ g(x,y)dy= 0. Such an equation is said to be "homogeneous" if and only if \(\displaystyle \frac{f(\lambda x, \lambda y)}{g(\lambda x, \lambda y)}= \frac{f(x, y)}{g(x, y)}\). Another way of saying that is that if we multiply both x and y by the same constant, the constant cancels out. In your equation, for example, \(\displaystyle f(x,y)= x^2+ y^2\) and \(\displaystyle g(x, y)= 2xy\) so that \(\displaystyle f(\lambda x, \lambda y)= (\lambda x)^2+ (\lambda y)^2= \lambda^2 x^2+ \lambda^2 y^2= \lambda^2(x^2+ y^2)\) and \(\displaystyle g(\lambda x, \lambda y)= 2(\lambda x)(\lambda y)= 2\lambda^2 xy\). So \(\displaystyle \frac{f(\lambda x,\lambda y)}{g(\lambda x, \lambda y)}= \frac{\lambda^2(x^2+ y^2)}{2\lambda^2 xy}= \frac{x^2+ y^2}{2xy}= \frac{f(x, y)}{g(x, y)}\).
The importance of homogeneous equations is that we can divide the entire equation by a power of x to get a separable equation with y/x rather than y as dependent variable. Here, dividing \(\displaystyle (x^2+ y^2)dx+ 2xydy= 0\) by \(\displaystyle x^2\) gives \(\displaystyle (1+ (y/x)^2)dx+ 2(y/x)dx\). Letting v= y/x, y= xv so dy= vdx+ xdv. The equation now is \(\displaystyle (1+ v^2)dx+ 2v(vdx+ xdv)= (1+ 3v^2)dx+ 2xvdv= 0\). That is separable and can be written \(\displaystyle \frac{dx}{x}= -\frac{2v}{1+ 3v^2}dv\). Integrate both sides.
The importance of homogeneous equations is that we can divide the entire equation by a power of x to get a separable equation with y/x rather than y as dependent variable. Here, dividing \(\displaystyle (x^2+ y^2)dx+ 2xydy= 0\) by \(\displaystyle x^2\) gives \(\displaystyle (1+ (y/x)^2)dx+ 2(y/x)dx\). Letting v= y/x, y= xv so dy= vdx+ xdv. The equation now is \(\displaystyle (1+ v^2)dx+ 2v(vdx+ xdv)= (1+ 3v^2)dx+ 2xvdv= 0\). That is separable and can be written \(\displaystyle \frac{dx}{x}= -\frac{2v}{1+ 3v^2}dv\). Integrate both sides.
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