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natima [27]
4 years ago
7

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

Mathematics
1 answer:
solniwko [45]4 years ago
5 0

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}}

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