Find the solution of $y'=10^{x-y}$p

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It is asked to find the solution of the ODE

$$y'=10^{x-y}$$

The given answer is $10^x+10^{-y}=C$, which I think is wrong. I thought ee should separate the variables and we have

$$\frac{dy}{dx}=\frac{10^x}{10^y} \Rightarrow 10^y dy = 10^xdx$$

Integrating:

$$\frac{10^y}{\log 10}=\frac{10^x}{\log 10} + k \Rightarrow y=c.x$$

this also doesnt seen right. What I did wrong?

Implicity,

$$10^x-10^y=C$$ seems right.

Thanks,

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2 Answers

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You went wrong in the very last step: $$ \frac{10^y}{\log 10} = \frac{10^y}{\log 10} + k \Longrightarrow y = \log_{10} \left( 10^x + k \right) $$ which is not at all the same as $ y = cx $.

You get the answer you thoght was right (and it is right) simply by multiplying through by $\log 10$.

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Up to here you are correct:$$\frac{10^y}{\log 10}=\frac{10^x}{\log 10}+k$$Now multiply both sides by $\log 10$ to get:$$10^y=10^x+k\log 10=10^x+C$$

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