Differential EquationsSolve the initial-value problem dydx=e2x/5y4,y(0)=4.y(x)=

Answer:[tex]y(x) = \sqrt[5]{\frac{e^{2x}}{2} + \frac{2047}{2}}[/tex]Step-by-step explanation:[tex]\displaystyle\frac{dy}{dx} = \displaystyle\frac{e^{2x}}{5y^4}[/tex]Cross multiplying, we have,[tex]5y^4 dy = e^{2x} dx[/tex]Integrating both sides,[tex]\int 5y^4 dy = \int e^{2x}dx[/tex]We obtain, [tex]y^5 = \frac{e^{2x}}{2} + C[/tex], where C is the constant of integration.[tex]y(x) = \sqrt[5]{\frac{e^{2x}}{2} + C }[/tex]We know that y(0) = 4Putting these value in the above equation, we get C = [tex]\frac{2047}{2}[/tex]Thus,[tex]y(x) = \sqrt[5]{\frac{e^{2x}}{2} + \frac{2047}{2}}[/tex]