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

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. !

• Please copy the form you want to the question. The link does not work. – robjohn Apr 5 '20 at 6:16
• You have a homogeneous and a inhomogeneous part. Work through in two separate cases. – Fakemistake Apr 5 '20 at 6:39

## 3 Answers

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.

• Where i am wrong Lutz Lehmann – jacky Apr 5 '20 at 6:41
• 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. – Lutz Lehmann Apr 5 '20 at 7:14

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.

• How can you apply "undetermined coefficients" when the coefficients of the linear DE are not constant? – Lutz Lehmann Apr 5 '20 at 6:55
• @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. – asd.123 Apr 5 '20 at 7:03
• In this the most trivial case it somehow all falls together, the inverse of the fundamental solution is the integrating factor. – Lutz Lehmann Apr 5 '20 at 7:09

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}