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Zepler [3.9K]
3 years ago
14

Solve the differential equation by variation of parameters. y''+y=secx

Mathematics
1 answer:
suter [353]3 years ago
8 0

Answer:y=

Acosx+Bsinx +cosx ln(cosx)+x sinx

Step-by-step explanation:

given  equation y''+y=secx

auxiliary  equation

p^2+1=0\\p=\pm i

so CF is y=Acosx+Bsinx

now

y_1(x)=cosx \ \ \ \ y_2(x)=sinx\\{y_1}'(x)=-sinx \ \ \ \ {y_2}'(x)=cosx

using wronskian formula

W=\begin{vmatrix}cosx &sinx \\ -sinx & cosx\end{vmatrix}

          =cos^2x+sin^2x=1

now f(x)=secx

u=-\int \frac{f(x)y_2(x)}{W(x)}dx \ and \ v=\int \frac{f(x)y_1(x)}{W(x)}dx

u=-\int \frac{secx\times sinx}{1}dx \ and \ v=\int \frac{secx\times cosx}{1}dx

u=-\int tanx dx \ and \ v=\int {1}dx

u=ln(cosx) \ \ \ and \ \ v=x

now particular integrals are

PI=cosx ln(cosx)+x sinx

total solution

y= C.F+P.I

y=Acosx+Bsinx +cosx ln(cosx)+x sinx

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Using the t-distribution to build the 99% confidence interval, it is found that:

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The confidence interval is:

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