# Find the general solution of dy/dt = t^3y.

*print*Print*list*Cite

### 4 Answers

The general solution of `dy/dt = t^3y` has to be determined.

`dy/dt = t^3y`

=> `(1/y) dy = t^3 dt`

take the integral of the two sides

`int (1/y) dy = int t^3 dt`

=> `ln y = t^4/4 + c`

=> `y = e^(t^4/4)*e^c`

To check the result, take the derivative, `dy/dt = e^c*e^(t^4/4)*t^3 = y*t^3`

**The required general solution of `dy/dt = t^3y` is **`y = e^(t^4/4)*e^c`

Given differential equation is `dy/dt=t^3y` .

Clearly the given differential equation is separable. So we take terms containing `y` with `dy` and terms containing `t` with `dt.`

We get `dy/y=t^3dt` .

Integrating both sides we get

`intdy/y=intt^3dt`

or, `lny=(1/4)t^4+lnc` , where `c` is the constant of integration.

or, `lny-lnc=(1/4)t^4`

or, `ln(y/c)=(1/4)t^4`

or, `y/c=e^((1/4)t^4)`

or, `y=ce^((1/4)t^4)` . This is the general solution of the given differential equation containing `c` as arbitrary constant of integration.

`(dy)/(dt)=t^3y`

`(dy)/(dt)-t^3y=0` (i) it is linear differential equation.

The integrating factor is `e^{int(-t^3)dt}=e^{-t^4/4}`

`` Thus (i) reduces to

`e^{-t^4/4}{(dy)/(dt)-t^3y}=e^(-t^4/3).0=0`

On integration ,we have

`ye^(-t^4/4)=c`

`` Ans.

its a differenzial equation:

`dy/dt=` `t^3 y`

separating y from t:

`(dy)/y= t^3 dt`

using integral:

`logy=1/4 t^4`

thus:

`y = e^(1/4 t^4) + c`

where c is a costant