Verify that $y=\mathrm{sin}x\mathrm{cos}x-\mathrm{cos}x$ is a solution of the initial value problem ${y}^{\prime}+(\mathrm{tan}x)y=co{s}^{2}xy(0)=-1$ on the interval $-\pi /2<x<\pi /2$

Caylee Villegas
2022-07-27
Answered

Verify that $y=\mathrm{sin}x\mathrm{cos}x-\mathrm{cos}x$ is a solution of the initial value problem ${y}^{\prime}+(\mathrm{tan}x)y=co{s}^{2}xy(0)=-1$ on the interval $-\pi /2<x<\pi /2$

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slapadabassyc

Answered 2022-07-28
Author has **21** answers

y'+y P(x)= Q(x)

${y}^{\prime}+(\mathrm{tan}x)y={\mathrm{cos}}^{2}x$ linear diff. eqn.

$P(x)=\mathrm{tan}x$

$Q(x)={\mathrm{cos}}^{2}x$

integration factor $i(x)=\mathrm{exp}(\int P(x)dx)=\mathrm{exp}(\int \mathrm{tan}(x)dx)$

$=\mathrm{exp}(-\mathrm{ln}(\mathrm{cos}x)=\frac{1}{\mathrm{cos}x}$

$y.i(x)=\int i(x)Q(x)dx+c$

$y\ast \frac{1}{\mathrm{cos}x}=\int \frac{1}{\mathrm{cos}x}{\mathrm{cos}}^{2}xdx+c$

$y\ast \frac{1}{\mathrm{cos}x}=\mathrm{sin}x+c$

$y=\mathrm{sin}x.\mathrm{cos}x+c.\mathrm{cos}x$

$y(0)=-1\Rightarrow -1=\mathrm{sin}0\mathrm{cos}0+c.cos0\Rightarrow c=-1$

$y=\mathrm{sin}x.\mathrm{cos}x-\mathrm{cos}x$

${y}^{\prime}+(\mathrm{tan}x)y={\mathrm{cos}}^{2}x$ linear diff. eqn.

$P(x)=\mathrm{tan}x$

$Q(x)={\mathrm{cos}}^{2}x$

integration factor $i(x)=\mathrm{exp}(\int P(x)dx)=\mathrm{exp}(\int \mathrm{tan}(x)dx)$

$=\mathrm{exp}(-\mathrm{ln}(\mathrm{cos}x)=\frac{1}{\mathrm{cos}x}$

$y.i(x)=\int i(x)Q(x)dx+c$

$y\ast \frac{1}{\mathrm{cos}x}=\int \frac{1}{\mathrm{cos}x}{\mathrm{cos}}^{2}xdx+c$

$y\ast \frac{1}{\mathrm{cos}x}=\mathrm{sin}x+c$

$y=\mathrm{sin}x.\mathrm{cos}x+c.\mathrm{cos}x$

$y(0)=-1\Rightarrow -1=\mathrm{sin}0\mathrm{cos}0+c.cos0\Rightarrow c=-1$

$y=\mathrm{sin}x.\mathrm{cos}x-\mathrm{cos}x$

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