# Solve differential equation (x+2)y′+4y=(3x + 6)^-2 lnx

waigaK 2021-02-24 Answered

Solve differential equation $$(x+2)y′+4y=(3x + 6)^-2 \ln x$$

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## Expert Answer

ottcomn
Answered 2021-02-25 Author has 21170 answers

$$((x+2)y'+4y)/((x+2))=((3x+6)^{-2} \ln x)/((x+2))$$
$$y'+(4/(x+2))y= \ln x/(9(x+2)^3)$$ (1)
Now, we know that solution of differential equation$$(y'+p(x)y= q(x))$$ is given by
$$y(I.F.)= \int q(x)(I.F.)dx$$ (2)
where $$I.F.= e^{\int p(x)dx}$$
So, finding integrating factor of equation (1)
$$I.F.= e^{\int 4/(x+2)dx}$$
$$= e^{4 \int dx/(x+2)}$$ $$( :' \int dx/(x+a)= \ln(x+a)+c)$$
$$= e^{4(\ln(x+2))}$$
$$= e^{\ln (x+2)^4}$$
$$= (x+2)^4$$
So, solution of given differential equation will be
$$y(x+2)^4= \int \ln x/(9(x+2)^3) (x+2)^4 dx$$
$$y(x+2)^4= 1/9 \int (x+2) \ln xdx$$
$$y(x+2)^4= 1/9 [\ln x \int(x+2)dx- \int 1/x [\int(x+2)dx]dx]$$
$$= 1/9 [((x+2)^2 \ln x)/2 - \int 1/x ((x+2)^2)/2 dx]$$
$$= ((x+2)^2 \ln x)/18-1/2 \int (x^2+4x+4)/x dx$$
$$= ((x+2)^2 \ln x)/18-1/2(x^2/2+4x+4\ln x)+c$$

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