Answer to Question #126307 in Differential Equations for jse

Question

Solve the given differential equation using an appropriate substitution. The DE is a bernoulli equation.

t^2 dy/dt + y^2 = ty

Answer 1

$t^2\frac{dy}{dt}+y^2=ty$

$t^2\frac{dy}{dt}-ty=-y^2$

Dividing equation by $-t^2y^2$ :

$-y^{-2}y'+y^{-1}t^{-1}=t^{-2}$

Substituting $z=y^{1-2}=y^{-1}$ , $\frac{dz}{dt}=z'=-y^{-2}y'$ , therefore: (1)

$z'+\frac z t =\frac 1 {t^2}$

Substituting $z=uv$ , $z'=u'v+uv'$ : (2)

$u'v+uv'+\frac {uv} t = \frac 1 {t^2}$

$u'v+u(v'+\frac {v} t) = \frac 1 {t^2}$ (3)

Setting the part inside () to 0 and separating variables:

$v'+\frac {v} t=0$

$\frac {dv}{dt}=-\frac v t$

$\int \frac {dv} v = - \int \frac {dt} t$

$ln(v) = -ln(t)+ln(k)$

$v=\frac k t$ (4)

Putting $v$ from (4) into equation (3):

$u'\frac k t = \frac 1 {t^2}$

$\int kdu = \int \frac {dt}t$

$u= \frac 1 k (ln(t)+C)$

Finding z from (2):

$z=uv=\frac 1k (ln(t)+C)\frac kt = \frac{ln(t)+C}{t}$

Finding y from (1):

$y=z^{-1}=\frac t {ln(t)+C}$

Answer: $y=\frac t {ln(t)+C}$