It follows that x(x+1)2x+1=x1+x+11 is the expression in partial fractions. Let us find the general solution of the differential equation x(x+1)dxdy=y(2x+1) expressing y explicitly in terms of x. We get that ydy=x(x+1)2x+1dx, and hence ∫ydy=∫x(x+1)2x+1dx=∫(x1+x+11)dx. We conclude that ln∣y∣=ln∣x∣+ln∣x+1∣+ln∣C∣=ln∣x(x+1)C∣. Therefore, y=Cx(x+1) is the general solution of the differential equation x(x+1)dxdy=y(2x+1).
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