Linear first-order differential equation with an initial condition

Lambda96
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Homework Statement
Show that the solution of a linear differential equation with boundary condition ##f(0)=f_0## has the following form ##f(x)=f_0 exp\biggl( \int_{0}^{x}ds ~g(s) \biggr)+\int_{0}^{x}ds ~h(s)exp\biggl( \int_{s}^{x}dr ~g(r)\biggr)##
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Hi,

unfortunately I have problems with the task d and e, the complete task is as follows:

Bildschirmfoto 2023-06-28 um 14.05.38.png

Bildschirmfoto 2023-06-28 um 14.06.04.png

I tried to form the derivative of the equation ##f(x)##, but unfortunately I have problems with the second part, which is why I only got the following.

$$\frac{d f(x)}{dx}=f_0 g(x) \ exp\biggl( \int_{0}^{x}ds \ g(s) \biggr)+?$$

I then tried another approach and first wanted to get rid of the integrals in the exp terms

$$f_0 exp\biggl( G(x)-G(0) \biggr)+\int_{0}^{x}ds \ h(s) exp \biggl( G(x)-G(s) \biggr) f(x)dx$$
$$f_0 e^{G(x)-G(0)}+e^{G(x)}\int_{0}^{x}ds \ h(s) e^{-G(s)}$$

Unfortunately, I now have problems again with the derivative for the second term using this method.
 
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You have an expression of the form <br /> \int_a^x f(x,t)\,dt for which you can show from the limit definition that, for sufficiently smooth f, <br /> \frac{d}{dx} \int_a^x f(x,t)\,dt = \lim_{h \to 0} \frac 1h \left( \int_a^{x+h} f(x+h, t)\,dt - \int_a^x f(x,t)\,dt\right) =<br /> f(x,x) + \int_a^x \left.\frac{\partial f}{\partial x}\right|_{(x,t)}\,dt.
 
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Likes DeBangis21 and Lambda96
Thanks pasmith for your help 👍 , I tried to implement your tip but am not sure about the ##f(x,x)## term:

I have now calculated the following

$$\frac{d f(x)}{dx}=\frac{d}{dx}\biggl( f_0 \ exp\Bigl( \int_{0}^{x}ds \ g(s) \Bigr) +\int_{0}^{x}ds \ h(s) \ exp\Bigl( \int_{s}^{x}dr \ g(r) \Bigr) \biggr)$$

$$\frac{d f(x)}{dx}= f_0 \ exp\Bigl( \int_{0}^{x}ds \ g(s) \Bigr) g(x) + h(x)exp\Bigl( \int_{s}^{x}dr \ g(r) \Bigr)+\int_{0}^{x}ds \ h(s) \frac{d}{dx} \ exp\Bigl( \int_{s}^{x}dr \ g(r) \Bigr)$$

$$\frac{d f(x)}{dx}= f_0 \ exp\Bigl( \int_{0}^{x}ds \ g(s) \Bigr) g(x) + h(x)exp\Bigl( \int_{s}^{x}dr \ g(r) \Bigr)+\int_{0}^{x}ds \ h(s) exp\Bigl( \int_{s}^{x}dr \ g(r) \Bigr)g(x)$$

$$\frac{d f(x)}{dx}= f_0 \ exp\Bigl( \int_{0}^{x}ds \ g(s) \Bigr) g(x) + h(x)exp\Bigl( \int_{s}^{x}dr \ g(r) \Bigr)+\int_{0}^{x}ds \ h(s) exp\Bigl( \int_{s}^{x}dr \ g(r) \Bigr)g(x)$$

$$\frac{d f(x)}{dx}= f_0 \ exp\Bigl( \int_{0}^{x}ds \ g(s) \Bigr) g(x) +\int_{0}^{x}ds \ h(s) exp\Bigl( \int_{s}^{x}dr \ g(r) \Bigr)g(x)+ h(x)exp\Bigl( \int_{s}^{x}dr \ g(r) \Bigr)$$

$$\frac{d f(x)}{dx}=f(x)g(x)+ h(x)exp\Bigl( \int_{s}^{x}dr \ g(r) \Bigr)$$

Now I'm just having trouble with the term ##h(x)exp\Bigl( \int_{s}^{x}dr \ g(r) \Bigr)##, which is equivalent to ##f(x,x)##. I was just having trouble figuring out what the exp term should be. After all, the easiest way would be for the integral to go from x to x, which would make the exponent zero and thus ##e^{0}=1## and I would get ##f(x)g(x)+h(x)##
 
Lambda96 said:
Now I'm just having trouble with the term ##h(x)exp\Bigl( \int_{s}^{x}dr \ g(r) \Bigr)##, which is equivalent to ##f(x,x)##. I was just having trouble figuring out what the exp term should be. After all, the easiest way would be for the integral to go from x to x, which would make the exponent zero and thus ##e^{0}=1## and I would get ##f(x)g(x)+h(x)##

Yes. The integrand is a function of x and s, so it is evaluated at (x,s) = (x,x).
 
Lambda96 said:
I then tried another approach and first wanted to get rid of the integrals in the exp terms

$$f_0 \exp\biggl( G(x)-G(0) \biggr)+\int_{0}^{x}ds \ h(s) \exp \biggl( G(x)-G(s) \biggr) \,dx$$
$$f_0 e^{G(x)-G(0)} + e^{G(x)} \int_{0}^{x}ds \ h(s) e^{-G(s)}$$

Unfortunately, I now have problems again with the derivative for the second term using this method.
The second term is of the form ##e^{G(x)} [J(x)-J(0)]## where ##J'(x) = h(x) \exp[-G(x)]##. Just apply the product rule.
 
Thanks pasmith and vela for your help 👍👍
 
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