Answer:
[tex]y(x)=\frac{2*(cos(4x)+2-4*C_1)}{cos(4x)-4C_1}[/tex]
Step-by-step explanation:
Rewrite the equation as:
[tex]\frac{dy(x)}{dx}=(y-2)^{2} *sin(4x)[/tex] (1)
divide both sides of (1) by [tex](y-2)^{2}[/tex]
[tex]\frac{\frac{dy}{dx} }{(y-2)^{2} } =sin(4x)[/tex]
Now integrate both sides:
[tex]\int\ \frac{1}{(y-2)^{2} } } \, dy = \int\ sin(4x) } } dx[/tex]
Solving the left side integral:
Let:
[tex]u=y-2\\du=dy[/tex]
Replacing [tex]u[/tex] and [tex]du[/tex]
[tex]\int\ \frac{du}{u^{2} } } }=-\frac{1}{u}[/tex]
[tex]u=y-2[/tex] then:
[tex]-\frac{1}{y-2}[/tex]
Solving the right side integral:
[tex]\int\ sin(4x) } } dx=-\frac{1}{4} cos(4x)+C_1[/tex]
Now we got this:
[tex]-\frac{1}{y-2}=-\frac{1}{4} cos(4x)+C_1[/tex]
Finally, solving for y:
[tex]y=\frac{2*(cos(4x)+2-4*C_1)}{cos(4x)-4C_1}[/tex]
