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Find that solution satisfying $ϕ(1) = 3ϕ(0)$ for the following second order linear ordinary differential equation:

$y' + 5y = 2$

I found the solution to be $ϕ(x) = \frac{2}{5} + ce^{5x}$.

Now how do I find a particular solution satisfying $ϕ(1) = 3ϕ(0)?$ Please help me with this. In the previous part, I was asked to find the solution satisfying $ϕ(1) = 2$ which I found to be $ϕ(x) = \frac{2}{5} + \frac{8}{5}e^5e^{-5x}$ if that helps.

3 Answers 3

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Just plug it in into the general form.

$$\phi(1)=\frac{2}{5}+ce^{-5}\overset{!}{=}\frac{6}{5}+3c = 3\phi(0)$$

This implies that

$$c\left(3-e^{-5}\right)=-\frac{4}{5}\implies c=-\frac{4}{5\left(3-e^{-5}\right)}$$

Plugging in, you have your solution as

$$\phi(x)=-\frac{4}{5\left(3-e^{-5}\right)}\cdot e^{-5x} + \frac{2}{5}$$

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Firstly it should be $$y'=2-5y\\ \frac { dy }{ 2-5y } =dx\\ \int { \frac { dy }{ 2-5y } } =\int { dx } \\ \int { \frac { d\left( 5y-2 \right) }{ 5y-2 } =-5\int { dx } } \\ \ln { \left| 5y-2 \right| =-5x+C } \\ 5y-2={ Ce }^{ -5x }\\$$

$$ \phi \left( x \right) =\frac { 2 }{ 5 } +{ Ce }^{ -5x }$$

then we are given $ϕ(1) = 3ϕ(0)$ so $$\frac { 2 }{ 5 } +C{ e }^{ -5 }=\frac { 6 }{ 5 } +3C\\ C\left( { e }^{ -5 }-3 \right) =\frac { 4 }{ 5 } \\ C=\frac { 4 }{ 5 } \left( { e }^{ -5 }-3 \right) $$

  • 0
    You are missing a factor of $\frac{1}{5}$ when doing the integral. – 2017-02-05
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    @Tom,thank you,it was a typo) – 2017-02-05
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Using integrating factor: $$e^{5x}(y'+5y) = 2 e^{5x},$$ $$\left(ye^{5x}\right)' = 2 e^{5x},$$ $$ye^{5x} = \frac25 e^{5x}+C,$$ $$y = \phi(x) = \frac25 + Ce^{-5x}.$$ Then: $$\phi(1) = 3\phi(0),$$ $$\frac25 + Ce^{-5} = 3\left(\frac25+C\right),$$ $$C=-\frac4{5(3-e^{-5})},$$ $$\phi(x)=\frac25-\frac{4}{5(3-e^{-5})}\,e^{-5x}$$