- #1
PhMichael
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Homework Statement
[tex]\int_{2}^{y(x)}e^{t^2+1}dt + \int_{x^2+3x}^{0}\frac{e^z}{1+z}=0[/tex]
I need so find [tex]y'(0)[/tex].
The Attempt at a Solution
[tex]\frac{d}{dx}\int_{2}^{y(x)}e^{t^2+1}dt =y'(x) \cdot e^{y(x)^2+1} [/tex]
[tex]\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)[/tex]
adding them and substituting x=0 yields:
[tex]y'(0) \cdot e^{y(0)^2+1} -3=0[/tex]
but how can i find y'(0) from them equation?