$\frac{dy}{dt}-y=(7e^t + 25e^{6t})$ in terms of y, when $y(0)=7$

Here's the problem:

$$\frac{dy}{dt}-y=7e^t + 25e^{6t}$$ in terms of $\,y\,$, when $\,y(0)=7\,$

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Please do check whether the above is what you meant. – DonAntonio Aug 31 '12 at 2:48
You should probably avoid the irrelevant comments in your question; it distracts from the mathematics. – Alex Becker Aug 31 '12 at 2:49
Thanks for the edit DonAntonio, and I believe that is the same problem. Also, I took out the irrelevant comments as per Alex Becker's request. – user13327 Aug 31 '12 at 2:52

To solve:

$$y'-y=7e^t+25e^{6t}$$

Multiply by $e^{-t}$:

$$e^{-t}y'-e^{-t}y=7+25e^{5t}$$

So that the right side is the result of the product rule, as follows:

$$(ye^{-t})'=7+25e^{5t}$$

Integrate both sides to find

$$ye^{-t}=7t+5e^{5t}+C$$ $$y=7te^t+5e^{6t}+Ce^t$$

Then just plug in the point $(t,y)=(0,7)$ given by your initial condition and solve for $C$. The resulting function for $y$ will be your answer. Though as a disclaimer I do have to point out that unless you have mathematically minded friends, this will probably impress them less than you'd like.

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Thank you very much :D – user13327 Aug 31 '12 at 2:58
Mathematically minded friends wouldn't be impressed by this :P – Tunococ Aug 31 '12 at 2:59

Put $y = a e^t + b te^t + c e^{6t}$ and solve for $a$, $b$, $c$.

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