Solve Example 4 using the general solution $y=a\sinh x+b\cosh x$.

The given general solution is $y=a\sinh x+b\cosh x$.

The correlation between the variables$x$and$y$obtained after removing the derivatives,where the relation includes arbitrary constants to represent the order of an equation, is the general solution of the differential equation. Aunique solution of the type$y=f\left(x\right)$that satisfies the differential equation is known as a particular solution. The general solution of the differential equation is used to derive the particular solution by giving values to the arbitrary constants.

First, find the second derivative of the general solution.

$\begin{array}{l}{y}^{\text{'}}=a\cosh x+b\sinh x\\ y^{\text{'}\text{'}}=a\sinh x+b\cosh x\end{array}$

Write the hyperbolic function in terms of exponential.

$\begin{array}{l}\sinh x=\frac{e^x-e^{-x}}{2}\\ \cosh x=\frac{e^x+e^{-x}}{2}\end{array}$

The general solution and its second derivative becomes

$\begin{array}{c}y=y^{\text{'}\text{'}}\\ =a\left(\frac{e^x-e^{-x}}{2}\right)+b\left(\frac{e^x+e^{-x}}{2}\right)\end{array}$

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Apply the boundary conditions $(0,0)$and $\left(\ln 2,\frac{3}{4}\right)$to find the particular solution.

$\begin{array}{c}0=b\\ \frac{3}{4}=a\left(\frac{e^{\ln 2}-e^{-\ln 2}}{2}\right)+0\\ =a\frac{3}{4}\\ =1\end{array}$

So, the particular solution is $y=\sinh x$.