Find the function y=y(x) (for x0) which satisfies the separable differential equation dy/dx=5+12x/xy^2; x0 with the initial condition y(1)=6 solve for y.

Question

Find the function y=y(x)  (for x>0) which satisfies the separable differential equation dy/dx=5+12x/xy^2; x>0 with the initial condition y(1)=6 solve for y.

myininaya · Answer

$$\frac{dy}{dx}=5+\frac{12x}{xy^2}$$
I don't think it can be done with separation of variables

anonymous · Answer

how do we solve then?

myininaya · Answer

well we can cancel an x in the second term so we have
$$y'=5+\frac{12}{y^2}$$
thinking...

myininaya · Answer

$$y^2y'=5y^2+12 => y^2y'-5y^2=12$$


hmm i can't think of how to do this one

anonymous · Answer

im not sure how to deal with it without the seperation of variables

myininaya · Answer

did you write the problem down correctly because you can't use separation on variables on this one

you could do it for this one:
$$\frac{dy}{dx}=\frac{5+12x}{xy^2}$$

but you wrote dy/dx=5+12x/xy^2 instead of dy/dx=(5+12x)/xy^2

anonymous · Answer

thats my mistake you are right

myininaya · Answer

ok goodie we can do this one then

myininaya · Answer

i will start by multiplying both sides by y^2
$$y^2 \frac{dy}{dx}=\frac{5+12x}{x}$$
now i will multiply dx on both sides
$$y^2 dy=\frac{5+12x}{x} dx$$
now integrate both sides
$$\frac{y^{2+1}}{2+1}+C_1=5\ln|x|+12x+C_2$$

myininaya · Answer

$$\frac{y^3}{3}=5\ln|x|+12x+C$$

myininaya · Answer

since $$C_2-C_1=constant $$

anonymous · Answer

solve for c?

myininaya · Answer

$$y^3=15\ln|x|+36x+C$$

myininaya · Answer

yes apply initial condition to find C

anonymous · Answer

C=180

anonymous · Answer

so if c=180 then $$y=\sqrt[3]{15\ln(x)+36x+180}$$

myininaya · Answer

6^3=36+c
36(6)=36+C
216=36+C
180=C
yes looks awesome :)

anonymous · Answer

thank you