Solution to DE Given Initial Condition and Value

[Sjsusu]

Hello, my calculus teacher recently assigned us these problems but he has not yet taught Differential Equations. I have been able to solve most of the other questions but I can't seem to figure this one out.
I have cross multiplied, integrated, and solved for c, but I can't find an expression that fits the initial condition to continue the problem.
Thanks for taking the time to read this, any help is much appreciated.

 

[MarkFL]

I would write:

[MATH]\int_0^{f(x)} e^y\,dy=\int_1^x e^u-1\,dx[/MATH]
Integrating, we obtain:

[MATH]e^{f(x)}-1=e^x-x-e+1[/MATH]
Can you proceed?

 

[Sjsusu]

Hello, I am still confused at how you integrated the problem. Where did the -1 come from after e^f(x) and -e+1 after e^x-x? Is it c?

 

[Sjsusu]

Hello, I am still confused at how you integrated the problem. Where did the -1 come from after e^f(x) and -e+1 after e^x-x? Is it c?

Oh nevermind I understand, I thought it was an indefinite integral. I am not sure what to do next though, but I think I need to take the ln of the equation sometime soon to single out f(x).

 

[MarkFL]

Oh nevermind I understand, I thought it was an indefinite integral. I am not sure what to do next though, but I think I need to take the ln of the equation sometime soon to single out f(x).

Yes, isolate \(e^{f(x)}\) and then convert to logarithmic form...what do you get?

 

[Sjsusu]

Yes, isolate \(e^{f(x)}\) and then convert to logarithmic form...what do you get?

B? After moving the -1 to the other side and taking the log I got about 0.349.

 

[MarkFL]

I get:

[MATH]f(x)=\ln(e^x-x-e+2)[/MATH]
Hence:

[MATH]f(-2)=\ln(e^{-2}-e+4)\approx0.34857968390496646\quad\checkmark[/MATH]

 

[Sjsusu]

Thank you!