d^2y/dt^2= 1-e^2t, y(1)=-1,y'(1)=0
What is the initial value?

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- anonymous

d^2y/dt^2= 1-e^2t, y(1)=-1,y'(1)=0
What is the initial value?

- schrodinger

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- myininaya

Do you mean solve the IVP?

- anonymous

Yes:)

- anonymous

The way I approach such a problem:
Solve the homogenous part of the differential equation first:
\[\Large y''(x)=0 \]
So the solution to this, almost trivial DE is:
\[\Large y_g(x)=Cx+K \]
Where C and K are constants of integration. Now you need to find the particular solution to the homogenous differential equation. This is usually done by guessing. Since:
\[\Large y''(x)=1-e^{2x} \]
A 'okay' guess could be:
\[\Large y_p(x)=A+Be^{2x} \]
Notice here that this is not yet good enough, see the solution already obtained above. We have Cx + K. So, in order to not overlap any solutions, we want to increase the linear exponent again:
\[\Large y_p(x)=Ax^2+Be^{2x} \]
Now this is a good guess. Substituting back will give you:
\[\Large 2A+4Be^{2x}=1-e^{2x} \]
From which you can obtain that:
\[ \Large A= \frac{1}{2}, \ B= -\frac{1}{4}\]
So given by the principle of superposition you have:
\[\Large y(x)=y_p(x)+y_g(x)=Cx+K+\frac{1}{2}x^2-\frac{1}{4}e^{2x} \]

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## More answers

- anonymous

I am not yet familiar with the methods you demonstrated above. Here is the answer I get: \[^{x^2/2}-^{e^2x/4}+^{e^2/4}-x-(1/2)\]

- anonymous

As I am new to typing equations, I wonder how you made your's look so neat?

- anonymous

I recommend you to check out Paul's website then, he has an amazing introduction to the method I suggested above, clearly, it's not the only one to solve this problem. But it makes inhomogeneous differential equations quite straight forward to solve.
Here is the link:
http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx
As to \(\LaTeX\).
First I use the perimeter \Large to make my equations look bigger, or in the case of this website - just more readable.
You can write fraction by typing \frac{nominator}{denominator}

- anonymous

Thanks for the help:)

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