solve this 1st order linear equation

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

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.

• $\frac{12+4x}{(x+3)^2}\neq 0.25x+0.75$. – vadim123 Sep 22 '18 at 16:56
• @vadim123 can you share what it is? I got that result through long division. – Nick Pavini Sep 22 '18 at 16:58

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