Question

How do you find all solutions of the differential equation $\dfrac{dy}{dx}=B+ky$ ?

Answer 1

To find all solutions of the differential equation $\dfrac{dy}{dx}=B+ky$ , we have to collect x terms including dy to one side and others to the other side. Then we will integrate both sides and find the value of y.

Complete step by step solution:
We have to find all solutions of the differential equation $\dfrac{dy}{dx}=B+ky$ . Let us find take the $B+ky$ to LHS and $dx$ to RHS.
$\Rightarrow \dfrac{dy}{B+ky}=dx$
Now, let us integrate both the sides. The above equation becomes
$\int{\dfrac{dy}{B+ky}}=\int{dx}...(i)$
Let us consider $u=B+ky$ . Let us differentiate both sides.
$\begin{aligned} & \Rightarrow du=kdy \\\ & \Rightarrow \dfrac{du}{k}=dy...(ii) \\\ \end{aligned}$
Now, let us multiply and divide LHS of equation (i) by k.
$\dfrac{1}{k}\int{\dfrac{kdy}{B+ky}}=\int{dx}$
Let us now substitute (ii) in the above equation.
$\dfrac{1}{k}\int{\dfrac{du}{u}}=\int{dx}$
We will now integrate. We know that $\int{\dfrac{1}{x}dx}=\log \left| x \right|+C$ and $\int{dx}=x$ . Hence, the above equation becomes
$\dfrac{1}{k}\log \left| u \right|=x+{{C}_{1}}$
Let us now substitute the value of u in the above equation.
$\Rightarrow \dfrac{1}{k}\log \left| B+ky \right|=x+{{C}_{1}}$
Now, we have to take l from LHS to RHS.
$\begin{aligned} & \Rightarrow \log \left| B+ky \right|=k\left( x+{{C}_{1}} \right) \\\ & \Rightarrow \log \left| B+ky \right|=kx+k{{C}_{1}} \\\ \end{aligned}$
Let us write $k{{C}_{1}}={{C}_{2}}$ since it’s a constant.
$\Rightarrow \log \left| B+ky \right|=kx+{{C}_{2}}$
We have to take anti-logarithm on both sides.
$\Rightarrow \left| B+ky \right|={{e}^{kx+{{C}_{2}}}}$
We know that ${{a}^{m+n}}={{a}^{m}}.{{a}^{n}}$ . Hence, the above equation becomes
$\Rightarrow \left| B+ky \right|={{e}^{kx}}{{e}^{{{C}_{2}}}}$
Let us denote ${{e}^{{{C}_{2}}}}$ as $C$ .
$\Rightarrow \left| B+ky \right|={{e}^{kx}}C$
Let us remove the absolute value. We are considering $B+ky$ to be positive.
$\Rightarrow B+ky={{e}^{kx}}C$
Let us take B to RHS.
$\Rightarrow ky={{e}^{kx}}C-B$
Now, we have to move k to RHS.
$\Rightarrow y=\dfrac{{{e}^{kx}}C-B}{k}$
Hence, the solution of differential equation $\dfrac{dy}{dx}=B+ky$ is $y=\dfrac{C{{e}^{kx}}-B}{k}$ , where $C={{e}^{k{{C}_{1}}}}$ .

Note: Students must know how to the rules and formulas of integration. We can also write the constants as C from the beginning. $\left| B+ky \right|$ denotes that $B+ky$ can also be negative, that is, $B+ky<0$ . But we usually take the positive condition.