Question #297064

<e> Solve the first order linear inhomogeneous differential equation using the bernoulli method

(2x+1)y,=4x+2y



1
Expert's answer
2022-02-25T10:30:29-0500

(2x+1)y=4x+2yy=u(x)v(x)(2x+1)(uv+uv)=4x+2uv(2x+1)uv+(2x+1)uv2uv=4x(2x+1)uv+u((2x+1)v2v)=4x1)(2x+1)v2v=0dvv=2dx2x+1dvv=2dx2x+1lnv=ln2x+1v=2x+12)(2x+1)u(2x+1)=4xdu=4x(2x+1)2u=4x(2x+1)2dx==2(2x+11)(2x+1)2dx==(22x+12(2x+1)2)dx==ln2x+1+12x+1+Cy=(2x+1)(ln2x+1+12x+1+C)(2x+1)y'=4x+2y\\ y=u(x)v(x)\\ (2x+1)(u'v+uv')=4x+2uv\\ (2x+1)u'v+(2x+1)uv'-2uv=4x\\ (2x+1)u'v+u((2x+1)v'-2v)=4x\\ 1) (2x+1)v'-2v=0\\ \frac{dv}{v}=\frac{2dx}{2x+1}\\ \int\frac{dv}{v}=\int\frac{2dx}{2x+1}\\ \ln|v|=\ln|2x+1|\\ v=2x+1\\ 2) (2x+1)u'\cdot (2x+1)=4x\\ du=\frac{4x}{(2x+1)^2}\\ u=\int\frac{4x}{(2x+1)^2}dx=\\ =\int\frac{2(2x+1-1)}{(2x+1)^2}dx=\\ =\int(\frac{2}{2x+1}-\frac{2}{(2x+1)^2})dx=\\ =\ln|2x+1|+\frac{1}{2x+1}+C\\ y= (2x+1)(\ln|2x+1|+\frac{1}{2x+1}+C)


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