Question

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

Answer 1

Answer:

$y(x) = \sqrt[5]{\frac{e^{2x}}{2} + \frac{2047}{2}}$

Step-by-step explanation:

$\displaystyle\frac{dy}{dx} = \displaystyle\frac{e^{2x}}{5y^4}$

Cross multiplying, we have,

$5y^4 dy = e^{2x} dx$

Integrating both sides,

$\int 5y^4 dy = \int e^{2x}dx$

We obtain,

$y^5 = \frac{e^{2x}}{2} + C$, where C is the constant of integration.

$y(x) = \sqrt[5]{\frac{e^{2x}}{2} + C }$

We know that y(0) = 4

Putting these value in the above equation, we get C = $\frac{2047}{2}$

Thus,

$y(x) = \sqrt[5]{\frac{e^{2x}}{2} + \frac{2047}{2}}$