Question 2.5.2: Find a family of solutions to the differential equation

First, we may write e^{t+2y} = e^{t} e^{2y} . Thus, we have

\frac {y´}{t}=e^{t} e^{2y}

Separating the variables, it follows that

e^{−2y} \frac {dy}{dt}= t e^{t}

ating both sides with respect to t , we may now write

\int {e^{−2y} dy}=\int {t e^{t} dt}

Using integration by parts on the right and evaluating both integrals, we have

−\frac {1}{2} e^{−2y} = (t −1)e^{t} +C

To now solve algebraically for y, we first multiply both sides by −2. Since C is an arbitrary constant, −2C is just another constant, one that we will denote by C_{1}. Hence

e^{−2y} = -2(t −1)e^{t} +C_{1}

Taking logarithms and solving for y, we can conclude that

y =−\frac {1}{2} ln(−2(t −1)e^{t} +C_{1})

is the family of functions that provides the general solution to the original DE.

To solve the corresponding IVP with y(1) = 1, we observe that

y(1)=−\frac {1}{2}ln(−2(1−1)e^{t} +C_{1})=−\frac {1}{2} ln(C_{1}) = 1

so ln(C_{1})=−2, and therefore C_{1} = e^{−2}. The solution to the IVP is

y =−\frac {1}{2}ln(−2(t −1)e^{1} +e^{−2})