Answer to Question #223119 in Discrete Mathematics for Iron rose

Question #223119

Express 2x+1 / x(x+1) in partial fractions and hence find the general solution of the differential equation x(x+1)dy/dx = y(2x+1) expressing y explicitly in terms of x.


1
Expert's answer
2021-09-15T02:57:08-0400

It follows that 2x+1x(x+1)=1x+1x+1\frac{2x+1}{x(x+1)}=\frac{1}{x}+\frac{1}{x+1} is the expression in partial fractions. Let us find the general solution of the differential equation x(x+1)dydx=y(2x+1)x(x+1)\frac{dy}{dx} = y(2x+1) expressing yy explicitly in terms of xx. We get that dyy=2x+1x(x+1)dx,\frac{dy}{y} = \frac{2x+1}{x(x+1)}dx, and hence dyy=2x+1x(x+1)dx=(1x+1x+1)dx.\int\frac{dy}{y} = \int\frac{2x+1}{x(x+1)}dx=\int(\frac{1}{x}+\frac{1}{x+1})dx. We conclude that lny=lnx+lnx+1+lnC=lnx(x+1)C.\ln|y|=\ln|x|+\ln|x+1|+\ln|C|=\ln|x(x+1)C|. Therefore, y=Cx(x+1)y=Cx(x+1) is the general solution of the differential equation x(x+1)dydx=y(2x+1)x(x+1)\frac{dy}{dx} = y(2x+1).


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