solve-the-initial-value-problem-d2ydx22e2xdisplaystyle-fracd2ydx2-2e2x-y03yprime

Question

Solve the initial value problem: d2ydx2=2e2x\displaystyle \frac{$$d^2y$$}{$$dx^2$$} = $$2e^{2x}$$, y′(0)=$$3y^{\prime}(0) = 3$$, y(0)=1y(0) = 1.

Accepted Answer

y(x)=12e2x+2x+12{\displaystyle\displaystyle y(x)=\frac{1}{2}$$e^{2x}+2x$$+\frac{1}{2}}

Answer

y(x)=e2x+2x+12{{\displaystyle{\displaystyle y(x)=$$e^{2x}+2x$$+\frac{1}{2}}}}

Answer

y(x)=12e2x+x+12{{\displaystyle{\displaystyle y(x)=\frac{1}{2}$$e^{2x}+x$$+\frac{1}{2}}}}

Answer

y(x)=12e2x+2x+1{{\displaystyle{\displaystyle y(x)=\frac{1}{2}$$e^{2x}+2x+1$$}}}

Solve the initial value problem: d2ydx2=2e2x\(\displaystyle\) \(\frac{d^2y}{dx^2}\) = 2e^{2x}, y′(0)=3y^{\(\prime\)}(0) = 3, y(0)=1y(0) = 1.