Question

How do you solve $y'+3y=0$ given $y\left( 0 \right)=4$?

Answer 1

In this problem, we have to solve the given differential equation and find the value of y. We can use the first order linear ordinary differential equation and we can derive an integrating factor and we can multiply the integrating factor to the differential equation. We can also solve this problem, by taking one term to the right-hand side and integrating on both sides and simply to solve for y.

Complete step by step answer:
We know that the given differential equation is,
$y'+3y=0$
We can write the term y’ as $\dfrac{dy}{dx}$, we get
$\dfrac{dy}{dx}+3y=0$
Now we can add -3y on both sides, we get
$\Rightarrow \dfrac{dy}{dx}=-3y$
Now we can rearrange the above step by separating the variables, that is treating $\dfrac{dy}{dx}$ as division and getting the y terms on one side and the x terms on the other side, we get
$\Rightarrow \dfrac{dy}{y}=-3dx$
We can now integrate on both the sides, we get

& \Rightarrow \int{\dfrac{dy}{y}}=\int{-3dx} \\\ & \Rightarrow \ln \left| y \right|=-3x+C\text{ }\because \int{\dfrac{1}{y}dy=\ln \left| y \right|},\int{\left( -3 \right)dx}=-3x \\\ \end{aligned}$$ Now we can use the initial condition $$y\left( 0 \right)=4$$, in the above step, we get $$\begin{aligned} & \Rightarrow \ln \left| 4 \right|=-3\left( 0 \right)+C \\\ & \Rightarrow C=\ln \left| 4 \right| \\\ \end{aligned}$$ We can substitute this C value, we get $$\Rightarrow \ln \left| y \right|=-3x+\ln \left( 4 \right)$$ Now we can take exponent on both the sides, we get $$\Rightarrow {{e}^{\ln \left| y\right|}}={{e}^{-3x+\ln \left| 4 \right|}}$$ Now we can use the exponent rule, we get $$\begin{aligned} & \Rightarrow y={{e}^{-3x}}.{{e}^{\ln \left| 4 \right|}} \\\ & \Rightarrow y={{e}^{-3x}}.4\text{ }\,\,\, \because {{\text{e}}^{\ln \left| 4 \right|}}=4 \\\ \end{aligned}$$ **Therefore, the value is $$y=4{{e}^{-3x}}$$.** **Note:** Students should know some basic integration formulas and rules to solve these types of problems. We should also know exponent rules to simplify this problem. We can also use the first order linear ordinary differential equation and we can derive an integrating factor and we can multiply the integrating factor to the differential equation to solve this problem.