Solve the initial value problem: `dy=e^(2x+y)dx ,y(0)=0.`

Explanation

`dy=e^(2x+y)dx `

`e^(-y)dy=e^(2x)dx`

Integrating,

`int e^(-y)dy=int e^(2x)dx`

`-e^(-y)=e^(2x)/2+C`

When `x=0,y=0`

So,

`-e^(0)=e^(0)/2+C`

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