Question

Question: In Problems , solve the equation.

$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{2}\mathbf{-}\sqrt{2x-y+3}$

Short answer

The solution of the given equation is $\mathbf{y}\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{3}\mathbf{-}\frac{{\left(x+C\right)}^2}{4}$.

Step-by-step solution

Given information and simplification

Given that, $\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{2}\mathbf{-}\sqrt{2x-y+3}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

Evaluate the equation (1).

Substitute u = 2x - y and $\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{2}\mathbf{-}\frac{\mathrm{du}}{\mathrm{dx}}$.

role="math" localid="1663945257707" $$\begin{gathered} \begin{array}{c}\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{2}\mathbf{-}\sqrt{2x-y+3}\\ \mathbf{2}\mathbf{-}\frac{\mathrm{du}}{\mathrm{dx}}\mathbf{=}\mathbf{2}\mathbf{-}\sqrt{u+3}\\ \frac{\mathrm{du}}{\mathrm{dx}}\mathbf{=}\sqrt{u+3}\\ \frac{1}{\sqrt{u+3}}\mathbf{du}\mathbf{=}\mathbf{dx}\end{array} \\ \frac{1}{\sqrt{u+3}}\mathbf{du}\mathbf{=}\mathbf{dx}\cdot \cdot \cdot \cdot \cdot \cdot \left(2\right) \end{gathered}$$

Now integrate the equation (2) on both sides.

$\int \frac{1}{\sqrt{u+3}}\mathbf{du}\mathbf{=}\int \mathbf{dx}$

Evaluation method

Find the value.

$\begin{array}{c}\int \frac{1}{\sqrt{u+3}}\mathbf{du}\mathbf{=}\int \mathbf{dx}\\ \mathbf{2}\sqrt{u+3}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{C}\\ \sqrt{u+3}\mathbf{=}\frac{x+C}{2}\\ \mathbf{u}\mathbf{+}\mathbf{3}\mathbf{=}{\left(\frac{x+C}{2}\right)}^2\end{array}$

Substitute the value of u here.

$\begin{array}{c}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{+}\mathbf{3}\mathbf{=}\frac{{\left(x+C\right)}^2}{4}\\ \mathbf{-}\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{3}\mathbf{+}\frac{{\left(x+C\right)}^2}{4}\\ \mathbf{y}\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{3}\mathbf{-}\frac{{\left(x+C\right)}^2}{4}\end{array}$

So, the solution is $\mathbf{y}\mathbf{=}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{3}\mathbf{-}\frac{{\left(x+C\right)}^2}{4}$