Question

Differential Equations

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

y(x)=

Answer 1

Answer:

$y^5\ =\ \dfrac{e^{2x}}{2}\ +\ \dfrac{7}{2}$

Step-by-step explanation:

Given,

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

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

Integrating on both sides, we have

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

$=>5\dfrac{y^5}{5}\ =\ \dfrac{e^{2x}}{2}\ +\ c$

$=>\ y^5\ =\ \dfrac{e^{2x}}{2}\ +\ c$

It is given that

y(o) = 4

$=>\ y^{(0)}\ =\ \dfrac{e^{2(0)}}{2}\ +\ c$

$=>\ 4\ =\ \dfrac{1}{2}\ +\ c$

$=>\ c\ =\ \dfrac{7}{2}$

Hence, the complete solution can be given by

$y^5\ =\ \dfrac{e^{2x}}{2}\ +\ \dfrac{7}{2}$