# `f'(s) = 10s - 12s^3, f(3) = 2` Find the particular solution that satisfies the differential equation.

### Textbook Question

Chapter 4, 4.1 - Problem 38 - Calculus of a Single Variable (10th Edition, Ron Larson).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to use direct integration to evaluate the general solution to the differential equation:

`int (10s - 12s^3)ds = int 10s ds - int 12s^3 ds`

`int (10s - 12s^3)ds = (10/2)s^2 - (12/4)s^4 + c`

`int (10s - 12s^3)ds = 5s^2 - 3s^4 + c`

You need to find the particular solution using the information provided by the problem, that f(3) = 2, such that:

`f(3) = 5*3^2 - 3*3^4 + c => 2 = 45 - 243 + c => c = 200`

Hence, evaluating the particular solution to the given differential equation yields `f(s) = 5s^2 - 3s^4 + 200.`

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To the differential function, you need to find the integral of f^1(s)

`int f^(s) = int f^1(s) ds`

= `int (10s - 12s^3) ds`

= `int 10s ds - int 12s^3 ds`

= `10 * (s^2)/ 2 - 12 *(s^4)/4 + C`

= `5s^2 - 3 s^4 + C`

You need to find the particular solution.It is given `f(3) = 2.`

`therefore f(3) = 5(3)^2 - 3(3)^4 + C = 2`

`= 45 - 243 + C = 2`

`therefore C = 200`

Hence particular differential solution is

`f(s) = 5s^2 - 3s^4 + 200`

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scisser | (Level 3) Honors

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Find the integral

`int(10s-12s^3)=f(s)=5s^2-3s^4+C`

Solve for C using the given point f(3)=2

`5(3)^2-3(3)^4+C=2 `

`C=200`

Thus,

`f(s)=5s^2-3s^4+200`