# differential equation

• Feb 9th 2010, 07:43 AM
JQ2009
differential equation
I have a 2nd order differential equation:

(d^2*y)/(d*x^2) = cosh(x)sinh(x)

How do you find a general solution?

Also how would you find a particular solution satisfying the conditions y(0) =1 and y'(0)= 2?

I really am having trouble understanding differential equations so help would be really appreciated.
• Feb 9th 2010, 08:38 AM
Nyrox
Recall that the general solution of a linear differential equation is the sum of a particular solution with the general solution of the corresponding homogeneous equation. The homogeneous equation has a trivial solution in this case:

$\displaystyle \frac{d^2y}{dx^2}=0 \Leftrightarrow y(x)=ax+b$ (with $\displaystyle a,b\in \mathbb R$)

As for the particular solution, notice that $\displaystyle \sinh$ and $\displaystyle \cosh$ are derivatives of one another. Thus, one could expect that a particular solution is given by $\displaystyle \alpha \sinh (x)\cosh (x)$ for some $\displaystyle \alpha \in \mathbb R$. Now simply observe that

$\displaystyle (\sinh(x)\cosh(x))''=(\sinh^2(x)+\cosh^2(x))'$
$\displaystyle =2\sinh(x)\cosh(x)+2\sinh(x)\cosh(x)=4\sinh(x)\cos h(x)$

so $\displaystyle \alpha =\frac{1}{4}$. The general solution is thus

$\displaystyle y(x)=\frac{1}{4}\sinh(x)\cosh(x)+ax+b$ (with $\displaystyle a,b\in \mathbb R$)