Respuesta :
Answer:
[tex]{y^2}=-\dfrac{cosx}{2}-\dfrac{cosx}{6}-\dfrac{1}{3}[/tex]
Step-by-step explanation:
Given that
[tex]y\ cosx\ {y}'=cos^3x\ sinx[/tex]
y(0)= -1
[tex]y\ cosx\ {y}'=cos^3x[/tex]
The above equation is a differential equation
So now by separating variables
[tex]y\ {y}'=cos^2x\ sinx[/tex]
We know that
[tex]cos^2x=1-sin^2x[/tex]
So
[tex]y\ {y}'=(1-1-sin^2x) sinx[/tex]
[tex]y\ {y}'=sinx-sin^3x [/tex]
[tex]ydy=\left (sinx-sin^3x \right )dx[/tex]
[tex]sin^3x=\dfrac{3sinx-sin3x}{4}[/tex]
Now by taking integration both side
[tex]\int ydy=\int \left (sinx-sin^3x \right )dx[/tex]
[tex]\int ydy=\int \left (sinx-\dfrac{3sinx-sin3x}{4} \right )dx[/tex]
[tex]\dfrac{y^2}{2}=-\dfrac{cosx}{4}-\dfrac{cosx}{12}+C[/tex]
Now by using y(0)= -1
[tex]C=-\dfrac{1}{6}[/tex]
[tex]\dfrac{y^2}{2}=-\dfrac{cosx}{4}-\dfrac{cosx}{12}-\dfrac{1}{6}[/tex]
[tex]{y^2}=-\dfrac{cosx}{2}-\dfrac{cosx}{6}-\dfrac{1}{3}[/tex]
