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

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

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Do you mean solve the IVP?

Yes:)

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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