How to Find y'(0) from an Integral Equation

[PhMichael]

Homework Statement

\int_{2}^{y(x)}e^{t^2+1}dt + \int_{x^2+3x}^{0}\frac{e^z}{1+z}=0

I need so find y'(0).

The Attempt at a Solution

\frac{d}{dx}\int_{2}^{y(x)}e^{t^2+1}dt =y'(x) \cdot e^{y(x)^2+1}

\frac{d}{dx}\int_{x^2+3x}^{0}\frac{e^z}{1+z}=-\frac{e^{x^2+3x}}{x^2+3x+1}\cdot (2x+3)

adding them and substituting x=0 yields:

y'(0) \cdot e^{y(0)^2+1} -3=0

but how can i find y'(0) from them equation?

 

[Quinzio]

Homework Statement

\int_{2}^{y(x)}e^{t^2+1}dt + \int_{x^2+3x}^{0}\frac{e^z}{1+z}=0

I need so find y'(0).

The Attempt at a Solution

\frac{d}{dx}\int_{2}^{y(x)}e^{t^2+1}dt =y'(x) \cdot e^{y(x)^2+1}

\frac{d}{dx}\int_{x^2+3x}^{0}\frac{e^z}{1+z}=-\frac{e^{x^2+3x}}{x^2+3x+1}\cdot (2x+3)

adding them and substituting x=0 yields:

y'(0) \cdot e^{y(0)^2+1} -3=0

but how can i find y'(0) from them equation?

y'(x)= \frac{d}{dx}\int_{2}^{y(x)}e^{t^2+1}dt + \frac{d}{dx}\int_{x^2+3x}^{0}\frac{e^z}{1+z}

Go with solving...

y'(0) = y'(0) \cdot e^{y(0)^2+1} -3

y'(0) =\frac{-3}{1-e^{y(0)^2+1}}

That's it.

 

[PhMichael]

but I need to find a number, not an expression involving y(0). how can y(0) be found?

 

[Quinzio]

You don't know y(x), so you can't know y'(0).