Respuesta :
[tex]y=C-3\cos[/tex]
[tex]y(\pi)=1[/tex]
[tex]\implies 1=C-3\cos\pi\implies 1=C+3\implies C=-2[/tex]
So the particular solution is
[tex]y=-2-3\cos x[/tex]
[tex]y(\pi)=1[/tex]
[tex]\implies 1=C-3\cos\pi\implies 1=C+3\implies C=-2[/tex]
So the particular solution is
[tex]y=-2-3\cos x[/tex]
Answer:
Hence, the particular solution is:
y= -2-3 cos(x)
Step-by-step explanation:
We are given a differential equation as:
y ′ = 3sin(x)
It's general solution is given as:
y = C −3cos(x) --------(1)
Now we are given a initial condition as:
y(π) = 1.
We will put the initial condition in the general solution to obtain the value of the constant 'C'.
We will put x=π in the equation (1).
y=C -3 cos(π)
As we know:
cos(π)=-1
Hence,
⇒ 1=C-3×(-1)
⇒ 1=C+3
⇒ C=1-3
⇒ C= -2
Hence, the particular solution is:
y= -2-3 cos(x)