Linear DE

[ehswhysee920]

(y^3)y'=(y^4)/(3t+y^5) , y(3/2)=1


How do I solve this DE? My professor told me that it is Linear but I'm confused because Linear DE's usually follow the pattern: y' + p(t)y = f(t) , I'm assuming that there is some way you'd manipulate this eqaution to make it fit that form, but I cant figure it out. Thank you.

 

[galactus]

It can be transformed.

Divide by y^3 and get:

\(\displaystyle \frac{dy}{dt}=\frac{y}{3t+y^{5}}\)

\(\displaystyle \frac{dt}{dy}=\frac{3t+y^{5}}{y}=\frac{3}{y}t+y^{4}\)

\(\displaystyle \frac{dt}{dy}-\frac{3}{t}y=y^{4}\)

Now, you can proceed and get:

\(\displaystyle \frac{y^{2}}{2}-\frac{t}{y^{3}}=C\)