Question

find initial value problem (ivp)
2y=5x+dy/dx y(0)=1

Answer 1

OK, so we're first going to put it in the standard form of y'+p(x)y=q(x): \[y'-2y=-5x\]We find an integrating factor, which will be \[u=e^{\int p(x)dx}=e^{-2x}\]and multiply both sides of the equation by it:
\begin{align}
u(y'-2y&)=u(-5x)\\
y' e^{-2x}-2ye^{-2x} &= -5xe^{-2x} \\
\left(ye^{-2x}\right)'&=-5xe^{-2x}\\
ye^{-2x}&=-5\int x e^{-2x}dx \\
ye^{-2x}&=-5\left(-\frac{1}{2}xe^{-2x}+\frac{1}{2}\int e^{-2x}dx\right)\\
ye^{-2x}&=-5\left(-\frac{1}{2}xe^{-2x}-\frac{1}{4}e^{-2x}+C\right)\\
ye^{-2x}&=\frac{5}{2}xe^{-2x}+\frac{5}{4}e^{-2x}+C\\
y&=\frac{5}{2}x+\frac{5}{4}+Ce^{2x}
\end{align}
Now we use the initial condition of y(0)=1 to solve for C:
\[1=\frac{5}{4}+Ce^{2\cdot 0}=\frac{5}{4}+C\]
Therefore C=-1/4 and your specific solution is \[y=\frac{5}{2}x+\frac{5}{4}-\frac{1}{4} e^{-2x}.\]