What is a solution to the differential equation $\frac{dy}{dx}=2y-1$?

Joyce Sharp
2022-09-23
Answered

What is a solution to the differential equation $\frac{dy}{dx}=2y-1$?

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Zackary Galloway

Answered 2022-09-24
Author has **17** answers

Substituting $z=2y-1$ in $\frac{dy}{dx}=2y-1$

and considering

$\frac{dz}{dx}=2\frac{dy}{dx}$ we have

$\frac{1}{2}\frac{dz}{dx}=z$. Grouping variables

$\frac{dz}{z}=2dx$ and integrating

$\mathrm{log}}_{e}z=2x+{C}_{1$ or equivalently

$z={C}_{2}{e}^{2x}=2y-1$ and finally

$y=\frac{{C}_{2}{e}^{2x}+1}{2}$

and considering

$\frac{dz}{dx}=2\frac{dy}{dx}$ we have

$\frac{1}{2}\frac{dz}{dx}=z$. Grouping variables

$\frac{dz}{z}=2dx$ and integrating

$\mathrm{log}}_{e}z=2x+{C}_{1$ or equivalently

$z={C}_{2}{e}^{2x}=2y-1$ and finally

$y=\frac{{C}_{2}{e}^{2x}+1}{2}$

asked 2022-05-15

Im clueless on how to solve the following question...

$x{e}^{y}\frac{dy}{dx}={e}^{y}+1$

What i've done is...

$\frac{dy}{dx}=\frac{1}{x}+\frac{1}{x{e}^{e}};\frac{dy}{dx}-\frac{1}{x{e}^{e}}=\frac{1}{x}$

Find the integrating factor..

$v(x)={e}^{P(x)};whereP(x)=\int p(x)dx\Rightarrow P(x)=\int \frac{1}{x}dx=ln|x|\phantom{\rule{0ex}{0ex}}v(x)={e}^{P(x)}={e}^{ln|x|}=x;\phantom{\rule{0ex}{0ex}}y=\frac{1}{v(x)}\int v(x)q(x)dx=\frac{1}{x}\int x\frac{1}{x}dx=1+c$

I know I made a mistake somewhere. Would someone advice me on this?

$x{e}^{y}\frac{dy}{dx}={e}^{y}+1$

What i've done is...

$\frac{dy}{dx}=\frac{1}{x}+\frac{1}{x{e}^{e}};\frac{dy}{dx}-\frac{1}{x{e}^{e}}=\frac{1}{x}$

Find the integrating factor..

$v(x)={e}^{P(x)};whereP(x)=\int p(x)dx\Rightarrow P(x)=\int \frac{1}{x}dx=ln|x|\phantom{\rule{0ex}{0ex}}v(x)={e}^{P(x)}={e}^{ln|x|}=x;\phantom{\rule{0ex}{0ex}}y=\frac{1}{v(x)}\int v(x)q(x)dx=\frac{1}{x}\int x\frac{1}{x}dx=1+c$

I know I made a mistake somewhere. Would someone advice me on this?

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I am having trouble isolating the x and y to separate side in the differential equations below. Could someone give me a hint as to how to to this.

Equation 1:

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Equation 2:

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Equation 1:

$\frac{dy}{dx}-\frac{x}{y}=\frac{1}{x}$

Equation 2:

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