Answer:
[tex]y=\sqrt[4]{2e^{2x}+4e^2x-1}[/tex]
Step-by-step explanation:
We begin with the differential equation [tex]\frac{dy}{dx} =\frac{e^{2x}}{y^3} +\frac{e^2}{y^3}[/tex]
First, we can factor out the [tex]\frac{1}{y^3}[/tex], separate the equation into a differential, and then integrate each side.
[tex]\frac{dy}{dx} =\frac{1}{y^3} (e^{2x}+e^2)\\\\\int y^3\, dy=\int(e^{2x}+e^2)dx\\\\\frac{1}{4} y^4=\frac{1}{2} e^{2x}+e^2x +C[/tex]
Now that we have integrated, we can apply our initial condition of [tex]y(0)=1[/tex] to find the value of C.
[tex]\frac{1}{4} (1)^4=\frac{1}{2} e^{2(0)}+e^2(0) +C\\\\\frac{1}{4} =\frac{1}{2} +C\\\\C=-\frac{1}{4}[/tex]
Now we can sub in C
[tex]\frac{1}{4} y^4=\frac{1}{2} e^{2x}+e^2x -\frac{1}{4}[/tex]
Now to solve for y
[tex]\frac{1}{4} y^4=\frac{1}{2} e^{2x}+e^2x -\frac{1}{4} \\\\y^4=2e^{2x}+4e^2x-1\\\\y=\sqrt[4]{2e^{2x}+4e^2x-1} \\y=-\sqrt[4]{2e^{2x}+4e^2x-1}[/tex]
As this has an even power, we have to decide whether to use the positive or negative version of the equation.
To test this, we can plug back in the Initial condition in order to see which one provides a correct solution
[tex]y=\sqrt[4]{2e^{2x}+4e^2x-1}\\\\1=\sqrt[4]{2e^{2(0)}+4e^2(0)-1}\\\\1=\sqrt[4]{2-1}\\\\1=1[/tex]
[tex]y=-\sqrt[4]{2e^{2x}+4e^2x-1}\\\\1=-\sqrt[4]{2e^{2(0)}+4e^2(0)-1}\\\\1=-\sqrt[4]{2-1}\\\\1=-1[/tex]
As we can see, the positive version is the only one that yields the correct solution, so this is our answer.
