Question

Dy/dx = (sin x)/y , y(0) = 2

Answer 1

The solution for this differential equation is
`y=√(-2cos(x)+6)`

Explanation:

This differential equation
`(dy)/(dx)=(sin(x))/(y)` is a separable First-Order ordinary differential equation.

We know this because a first-order differential equation is separable if and only if it can be written as

`(dy)/(dx)=f(x)g(y)` where f and g are known functions.

And we have

`(dy)/(dx)=(sin(x))/(y)\\ (dy)/(dx)=sin(x)(1)/(y)`

To solve this differential equation we need to integrate both sides

`y\cdot dy=sin(x)\cdot dx\\ \int\limits {y\cdot dy}= \int\limits {sin(x)\cdot dx}`

`\int\limits {y\cdot dy}=(y^(2) )/(2) + C`

`\int\limits {sin(x) \cdot dx}=-cos(x) + C`

`(y^(2) )/(2) + C=-cos(x) + C`

We can make a new constant of integration
`C_(1)`

`(y^(2) )/(2)=-cos(x) + C_(1)`

We need to isolate y

`(y^(2) )/(2)=-cos(x) + C_(1)\\y^2=-2cos(x)+2*C_(1)\\\mathrm{For\:}y^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}y=√(f\left(a\right)),\:\:-√(f\left(a\right))\\y=\sqrt{-2cos(x)+c_(1) } \\y=-\sqrt{-2cos(x)+c_(1) }`

We have the initial conditions y(0)=2 so we can find the value of the constant of integration for
`y=\sqrt{-2cos(x)+c_(1) }`

`2=√(-2\cos \left(0\right)+c_1)\\2= √(-2+c_1) \\c_1=6`

For
`y=-\sqrt{-2cos(x)+c_(1) }` there is not solution for
`c_(1)` in the domain of real numbers.

The solution for this differential equation is
`y=√(-2cos(x)+6)`