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How can i solve the differential equation

$$\frac{dy}{dx}-2(3\cos x+5)y=-1$$

What i have try

It represent a linear differential equation of degree and order $1$

So compare with $\frac{dy}{dx}+Py=Q$

We have $P=-2(3\cos x+5)$ and $Q=-1$

And Integrating factor $\text{(I.f)} =e^{\int 2(3\cos x+5)dx}=e^{-2(3\sin x+5x)}$

So solution is $$ y=\int Q\text{(I.f)}dx=-\int e^{-2(3\sin x+5x)}dx$$

How do i solve it Help me please or How can i write its solution . Thanks

Update: wolframalpham alpha show as

How can i write it solution in that form.

enter image description here!

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    $\begingroup$ Please copy the form you want to the question. The link does not work. $\endgroup$ – robjohn Apr 5 '20 at 6:16
  • $\begingroup$ You have a homogeneous and a inhomogeneous part. Work through in two separate cases. $\endgroup$ – Fakemistake Apr 5 '20 at 6:39
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You have somehow missed a part of the solution formula. With the integrating factor $\mu(x)=\exp(\int P(x) dx)$ you get $$ (\mu(x) y(x))'=\mu (x)y'(x)+\mu'(x) y(x)=\mu(x)(y'(x)+P(x)y(x))=\mu(x) Q(x) \\~\\ \implies \mu(x)y(x)=\int \mu(x)Q(x)\,dx $$ which means that in the formula for $y$ you also have to divide by $\mu$. Together with the integration constant that is hidden in the undetermined integral, you get exactly the same solution formula as WA.

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  • $\begingroup$ Where i am wrong Lutz Lehmann $\endgroup$ – jacky Apr 5 '20 at 6:41
  • $\begingroup$ The solution formula for the integrating factor is $y(x)=\frac1{\mu(x)}\int\mu(x)Q(x)dx$, your formula is missing the first factor. $\endgroup$ – Lutz Lehmann Apr 5 '20 at 7:14
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I started from: \begin{equation} e^{\int P(x) d x}\left(\frac{d y}{d x}+P(x) y\right)=Q(x) e^{\int P(x) d x} \end{equation} Then: \begin{equation} \begin{array}{c} e^{\int P(x) d x}\left(\frac{d y}{d x}+P(x) y\right)=\frac{d}{d x}\left(e^{\int P(x) d x} y\right) \\ \frac{d}{d x}\left(e^{\int P(x) d x} y\right)=Q(x) e^{\int P(x) d x} \end{array} \end{equation} Integrate both sides of the new equation: \begin{equation} \int \frac{d}{d x}\left(e^{\int P(x) d x} y\right) d x=\int Q(x) e^{\int P(x) d x} d x \end{equation} The Fundamental Theorem of Calculus shows that: \begin{equation} \int \frac{d}{d x}\left(e^{\int P(x) d x} y\right) d x=e^{\int P(x) d x} y+C_{1} \end{equation} where C1 is an arbitrary constant and RHS also needs to be found and let it B(x)+ C2, where C2 is a constant due to the integral. \begin{equation} e^{\int P(x) d x} y=B(x)+C_{3} \end{equation} Divide by the integrating factor to get the solution: \begin{equation} y=B(x) e^{-\int P(x) d x}+C_{3} e^{-\int P(x) d x} \end{equation} Basically repeating the order will lead you to the solution.

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    $\begingroup$ How can you apply "undetermined coefficients" when the coefficients of the linear DE are not constant? $\endgroup$ – Lutz Lehmann Apr 5 '20 at 6:55
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    $\begingroup$ @LutzLehmann Yes correct I want to write variations of parameter it is more general but somehow I write there undetermined coefficients, sorry for that I will edit. $\endgroup$ – asd.123 Apr 5 '20 at 7:03
  • $\begingroup$ In this the most trivial case it somehow all falls together, the inverse of the fundamental solution is the integrating factor. $\endgroup$ – Lutz Lehmann Apr 5 '20 at 7:09
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Homogeneous Part $$y'(x)-\underbrace{2(3\cos(x)+5)}_{p(x)}\;y(x)=0$$ Solution: $y(x)=c\exp[P(x)]$ with $P'(x)=p(x)$ and $c\in\mathbb{R}$. In particular $$P(x)=\int p(x)\,\mathrm{d}x=\int [6\cos(x)+10]\, \mathrm{d}x=6\sin{x}+10x$$ Inhomogeneous Part $$y'(x)-\underbrace{2(3\cos(x)+5)}_{p(x)}y(x)=-1$$ Variation of the constant: $y(x)=c(x)\;\underbrace{\exp[6\sin(x)+10x]}_{h(x)}$. Note that $h(x)$ is a solution of the homogeneous part, thus when you plug into the original equation $$c'(x)h(x)+c(x)h'(x)-p(x)c(x)h(x)=-1$$ Since $h'(x)=p(x)h(x)$, you get $$c'(x)=-\frac{1}{h(x)}=-1\exp[-(6\sin(x)+10x)]$$ Integrate: $$c(x)=c+\int -1\exp[-(6\sin(x)+10x)]\,\mathrm{d}x$$ Put them all together: \begin{align} y(x)&=\Big(c+\int -1\exp[-(6\sin(x)+10x)]\,\mathrm{d}x\Big)\exp[6\sin (x)+10x]\\ &=ce^{6\sin(x)+10x}+e^{6\sin(x)+10x}\int e^{-(6\sin(x)+10x)}\,\mathrm{d}x \end{align}

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