Question

Solve the initial-value problem

y cos x y' = (cosx)3 sinx, y(0) = −1.

Note: The cosx is raised to a power of 3: (cosx)^3

Answer 1

`{y^2}=-(cosx)/(2)-(cosx)/(6)-(1)/(3)`

Explanation:

Given that

`y\ cosx\ {y}'=cos^3x\ sinx`

y(0)= -1

`y\ cosx\ {y}'=cos^3x`

The above equation is a differential equation

So now by separating variables

`y\ {y}'=cos^2x\ sinx`

We know that

`cos^2x=1-sin^2x`

So

`y\ {y}'=(1-1-sin^2x) sinx`

`y\ {y}'=sinx-sin^3x`

`ydy=\left (sinx-sin^3x \right )dx`

`sin^3x=(3sinx-sin3x)/(4)`

Now by taking integration both side

`\int ydy=\int \left (sinx-sin^3x \right )dx`

`\int ydy=\int \left (sinx-(3sinx-sin3x)/(4) \right )dx`

`(y^2)/(2)=-(cosx)/(4)-(cosx)/(12)+C`

Now by using y(0)= -1

`C=-(1)/(6)`

`(y^2)/(2)=-(cosx)/(4)-(cosx)/(12)-(1)/(6)`

`{y^2}=-(cosx)/(2)-(cosx)/(6)-(1)/(3)`