\frac{dy}{dx}+5y=15

$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{5}{y}={15}$$

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Mollie Nash
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{5}{y}={15}$$
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={15}-{5}{y}$$
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{15}-{5}{y}}}}={\left.{d}{x}\right.}\Rightarrow{\frac{{{\left.{d}{y}\right.}}}{{{5}{\left({3}-{y}\right)}}}}={\left.{d}{x}\right.}$$
$$\displaystyle\Rightarrow\int{\frac{{{\left.{d}{y}\right.}}}{{{3}-{y}}}}=\int{5}{\left.{d}{x}\right.}$$
$$\displaystyle\Rightarrow-{\log{{\left({3}-{y}\right)}}}={5}{x}+{\log{{c}}}$$
$$\displaystyle\Rightarrow-{5}{x}={\log{{\left({3}-{y}\right)}}}+{\log{{c}}}$$
$$\displaystyle={\log{{c}}}{\left({3}-{y}\right)}$$
$$\displaystyle\Rightarrow{c}{\left({3}-{y}\right)}={e}^{{-{5}{x}}}$$
$$\displaystyle{3}-{y}={\frac{{{e}^{{-{5}{x}}}}}{{{c}}}}\Rightarrow{y}={3}-{\frac{{{e}^{{-{5}{x}}}}}{{{c}}}}$$
$$\displaystyle{y}={3}-{\frac{{{e}^{{-{5}{x}}}}}{{{c}}}}$$
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alkaholikd9
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{5}{y}={15}$$
we use variable seperable method subtract both sides by 5y
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={15}-{5}{y}$$
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}=-{5}{\left({y}-{3}\right)}$$
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{y}-{3}}}}=-{5}{\left.{d}{x}\right.}$$
$$\displaystyle\int{\frac{{{\left.{d}{y}\right.}}}{{{y}-{3}}}}=\int-{5}{\left.{d}{x}\right.}$$
$$\displaystyle{\ln{{\left({y}-{3}\right)}}}=-{5}{x}+{C}$$
take exponent both sides
$$\displaystyle{e}^{{{\ln{{\left({y}-{3}\right)}}}}}={e}^{{-{5}{x}+{C}}}$$
$$\displaystyle{y}-{3}={C}{e}^{{-{5}{x}}}$$
Answer: $$\displaystyle{y}={3}+{C}{e}^{{-{5}{x}}}$$
karton

$$\begin{array}{}\frac{dy}{dx}=15-5y \\\frac{dy}{dx}=-5(y-3) \\\frac{dy}{y-3}=-5dx \\\int \frac{dy}{y-3}=\int -5dx \\\ln (y-3)=-5x+C \\e^{\ln (y-3)}=e^{-5x+C} \\y-3=Ce^{-5x} \\y=3+Ce^{-5x} \end{array}$$

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