Question

Please help explain this

Answer 1

(a) -
`\sin(y)=-3\cos(x)+4`

(b) -
`y=e^(2x-1)`

Explanation:

Given the two part question...

(a) - Solve the following first-order differential equation with the given initial condition.

`y'=(3\sin(x))/(\cos(y)); \ y(0)=(\pi)/(2)`

(b) - Given that y=f(x) passes through the point (1,e) and has the property that the slope of the tangent line to the graph at any point "P" is equal to twice the y-coordinate of "P." Find f(x).

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Part (a) -

The given DE is a separable differential equation and can be solved in the following manner.

`\boxed{\left\begin{array}{ccc}\text{\underline{Seperable Differential Equation}}\\(dy)/(dx)=f(x)g(y) \\ \rightarrow \int(dy)/(g(y))=\int f(x)dx \end{array}\right}`

`y'=(3\sin(x))/(\cos(y)) \rightarrow (dy)/(dx) =(3\sin(x))/(\cos(y))\\\\\Longrightarrow \cos(y)dy=3\sin(x)dx\\\\\Longrightarrow \int\cos(y)dy=3\int\sin(x)dx\\\\\Longrightarrow \boxed{\sin(y)=-3\cos(x)+C}\\\\\text{Now for the given initial condition to find the arbitrary constant,`

`\therefore \boxed{\boxed{\sin(y)=-3\cos(x)+4}}`

Thus, the DE is solved.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Part (b) -

`\text{The slope of the tangent line =} \ (dy)/(dx) \\\\\text{According to the question, the slope of the tangent line =} \ 2y\\\\\text{So we can say,}\\\\\longrightarrow \boxed{(dy)/(dx)=2y} \ \text{Solve the DE}`

`(dy)/(dx)=2y\\\\\Longrightarrow (dy)/(2y)=dx \\\\\Longrightarrow (1)/(2) \int(dy)/(y)=\int dx\\\\\Longrightarrow (1)/(2) \ln(y)=x+C\\\\\Longrightarrow \boxed{\ln(y)=2x+ \bar C}`

Use the given initial condition to find "C." y(e)=1

`\Longrightarrow \ln(e)=2(1)+ \bar C\\\\\Longrightarrow 1=2+ \bar C\\\\\Longrightarrow \boxed{C=-1}`

`\Longrightarrow \ln(y)=2x+ \bar C\\\\\Longrightarrow \ln(y)=2x-1\\\\ \therefore \boxed{\boxed{ y=e^(2x-1)}}`

Thus, f(x) is found.