Question

Por favor! solução geral da equação diferencial: dy/dx = y+3/(2x−5)^6

Answer 1

$Sabemos\ que: \\\\ y'= \frac{y+3}{(2x-5)^6}\ --\ \textgreater \ \ y'-\frac{y}{(2x-5)^6}=\frac{3}{(2x=5)^6}\\\\ y'+p(x).y=q(x)\ --\ \textgreater \ \ p(x)=\frac{-1}{(2x-5)^6}\\\\ q(x)=\frac{3}{(2x-5)^6}\\\\\\ Calculando\ o\ fator\ integrante:\\\\ \mu'(x)=\mu(x).p(x)\ --\ \textgreater \ \mu'(x)=\mu(x).(\frac{-1}{(2x-5)^6})\\\\ \frac{\mu'(x)}{\mu(x)}= \frac{\mu(x).(\frac{-1}{(2x-5)^6})}{\mu(x)}\ --\ \textgreater \ \ \frac{d}{dx}(ln(\mu(x))= \frac{-1}{(2x-5)^6}\\\\ ln(\mu(x))=\int {\frac{-1}{(2x-5)^6}}.dx$

$Resolvendo\ \int \frac{-1}{(2x-5)^6}}.dx:\\\\ u=2x-5\ \|\ \frac{du}{2}=dx\\\\ \int\frac{-1}{u^6}.\frac{1}{2}.du\ --\ \textgreater \ \ \frac{-1}{2}\int u^{-6}.du\\\\ \frac{-1}{2}.\frac{u^{-5}}{-5}\ --\ \textgreater \ \ \frac{1}{10.u^5}\ --\ \textgreater \ \ \frac{1}{10.(2x-5)^5}+c\\\\ Logo:\\\\ ln(\mu(x))= \frac{1}{10.(2x-5)^5}\ --\ \textgreater \ \ \mu(x)=e^{\frac{1}{10.(2x-5)^5}$

$Voltando\ para\ a\ EDO:\\\\ \frac{d}{dx}(\mu(x).y)=\mu(x).q(x)\\\\ \frac{d}{dx}(e^{\frac{1}{10(2x-5)^5}}.y)=e^{\frac{1}{10(2x-5)^5}}.\frac{3}{(2x-5)^6}\\\\ e^{\frac{1}{10(2x-5)^5}}.y=\int \frac{3.e^{\frac{1}{10(2x-5)^5}}}{(2x-5)^6}$

$Resolvendo\ \int \frac{3.e^{\frac{1}{10(2x-5)^5}}}{(2x-5)^6}:\\\\ u=\frac{1}{10.(2x-5)^5}\\\\ \frac{du}{dx}= \frac{d}{dx}(\frac{1}{10.(2x-5)^5})= \frac{1}{10}.\frac{d}{dx}((2x-5)^{-5})\\\\ \frac{1}{10}.(-5.(2x-5)^{-6}.2)=\frac{1}{10}.(-10).(2x-5)^{-6}\\\\ \frac{du}{dx}=\frac{-1}{(2x-5)^6}\ --\ \textgreater \ \ du=\frac{-1}{(2x-5)^6}.dx\\\\ \int 3.e^u.(-1).du\ --\ \textgreater \ \ -3\int e^u.du\\\\ -3.e^u=-3.e^{\frac{1}{10.(2x-5)^5}}+c$

$Voltando\ para\ a\ EDO:\\\\ e^{\frac{1}{10(2x-5)^5}}.y=-3.e^{\frac{1}{10(2x-5)^5}}+c\\\\ y= \frac{-3.e^{\frac{1}{10(2x-5)^5}}+\ c}{e^{\frac{1}{10(2x-5)^5}}}\\\\\\ Expandindo\ (2x-5)^5:\\\\ (2x-5)^5=C_{5;0}.(2x)^5.(-5)^0+C_{5;1}.(2x)^4.(-5)^1+C_{5;2}.(2x)^3.(-5)^2\\ +C_{5;3}.(2x)^2.(-5)^3+C_{5;4}.(2x)^1.(-5)^4+C_{5;5}.(2x)^0.(-5)^5\\\\ (2x-5)^5=32x^5+5.16x^4.(-5)+10.8x^3.25+10.4x^2.(-125)+\\ +5.2x.625-3125\\\\ (2x-5)^5=32x^5-400x^4+2000x^3-5000x^2+6250x-3125$

$Logo:\\\\ y= \frac{-3.e^{\frac{1}{10.(32x^5-400x^4+2000x^3-5000x^2+6250x-3125)}}+\ c}{e^{\frac{1}{10.(32x^5-400x^4+2000x^3-5000x^2+6250x-3125)}}} \\\\\\ \| \ y=\frac{-3.e^{\frac{1}{320x^5-4000x^4+20000x^3-50000x^2+62500x-31250}}+\ c}{e^{\frac{1}{320x^5-4000x^4+20000x^3-50000x^2+62500x-31250}}}\ \|$