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Mathematics

1 answer:

artcher [175]3 years ago

5 0

Answer:

1,5

Step-by-step explanation:

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Solve this problem by using variation of parameters method.<br> y''-y=coshx.

Gekata [30.6K]

$y''-y=0\implies r^2-1=0\implies r=\pm1$
$\implies y_c=C_1e^x+C_2e^{-x}={C^*}_1\underbrace{\cosh x}_{y_1}+{C^*}_2\underbrace{\sinh x}_{y_2}$

For the nonhomogeneous ODE

$y''-y=\underbrace{\cosh x}_{f(x)}$

we're looking for a particular solution of the form

$y_p=u_1y_1+u_2y_2$

where

$u_1=-\displaystyle\int\frac{y_2(x)f(x)}{W(y_1(x),y_2(x))}\,\mathrm dx$
$u_2=\displaystyle\int\frac{y_1(x)f(x)}{W(y_1(x),y_2(x))}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the two fundamental solutions.

We have

$W(y_1,y_2)=\begin{vmatrix}\cosh x&\sinh x\\\sinh x&\cosh x\end{vmatrix}=\cosh^2x-\sinh^2x=1$

so we're left with

$u_1=-\displaystyle\int\sinh x\cosh x\,\mathrm dx=-\dfrac12\cosh^2x$
$u_2=\displaystyle\int\cosh^2x\,\mathrm dx=\dfrac12x+\dfrac14\sinh2x$

so that the particular solution is

$y_p=-\dfrac12\cosh^3x+\dfrac12x\sinh x+\dfrac14\sinh x\sinh2x$
$y_p=-\dfrac12\cosh x+\dfrac12x\sinh x$

As $y_1$ already accounts for the $\cosh x$ term in $y_p$ , we're left with the general solution

$y=C_1\cosh x+C_2\sinh x+\dfrac12x\sinh x$