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$$(x+3)^2\frac{dy}{dx}=6-12y-4xy=6-y(12+4x)$$

a. Write it in standard form. $$\frac{dy}{dx}+\frac{12+4x}{(x+3)^2}y=\frac6{(x+3)^2}$$

b. What is the integrating factor? $$\frac{12+4x}{(x+3)^2} = \frac14x+\frac34 \implies$$ $$IF=e^{\int{\frac14x+\frac34}}=e^{\frac18x^2+\frac34x}$$

c. Integrate the DE subject to $y(0)=1$. $$e^{\frac18x^2+\frac34x}\frac{dy}{dx}+ye^{\frac18x^2+\frac34x}(\frac14x+\frac34)=\frac{6e^{\frac18x^2+\frac34x}}{(x+3)^2} \implies$$ $$ye^{\frac18x^2+\frac34x}=\int{\frac{6e^{\frac18x^2+\frac34x}}{(x+3)^2}}+c$$

I have done this problem a few times and now getting kind of stuck. Please check my standard form and integrating factor.

Right now if everything else is correct I cannot solve the integral. Any help appreciated.

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  • $\begingroup$ $\frac{12+4x}{(x+3)^2}\neq 0.25x+0.75$. $\endgroup$ – vadim123 Sep 22 '18 at 16:56
  • $\begingroup$ @vadim123 can you share what it is? I got that result through long division. $\endgroup$ – Nick Pavini Sep 22 '18 at 16:58
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$$(x+3)^2\frac{dy}{dx}=6-4y(3+x)$$ $$(x+3)^2\frac{dy}{dx}+4y(3+x)=6$$ Multiply by $(x+3)^2$ $$(x+3)^4y'+4y(3+x)^3=6(x+3)^2$$ $$((x+3)^4y)'=6(x+3)^2$$ Integrate $$(x+3)^4y=2(x+3)^3+C$$ $$y(x)=\frac 2 {(x+3)}+\frac C {(x+3)^4}$$


As pointed out in the comment this line is not correct $$\frac{12+4x}{(x+3)^2} = \frac14x+\frac34 $$ This is correct $$\frac{12+4x}{(x+3)^2} = \frac 4 {x+3}$$

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$$\frac{12+4x}{(x+3)^2}=\frac{4(x+3)}{(x+3)^2}=\frac{4}{x+3}$$

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  • $\begingroup$ Wow. Tunnel vision really got to me on this one. Thank you. $\endgroup$ – Nick Pavini Sep 22 '18 at 17:01
  • $\begingroup$ Glad to be of service. $\endgroup$ – vadim123 Sep 22 '18 at 17:04

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