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

justaguide | College Teacher | (Level 2) Distinguished Educator

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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`

rakesh05 | High School Teacher | (Level 1) Assistant Educator

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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.

pramodpandey | College Teacher | (Level 3) Valedictorian

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`(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.

oldnick | (Level 1) Valedictorian

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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