Question

In Problems 3–6 the dependent variable y is missing in the given differential equation. Proceed as in Example 1 and solve the equation by using the substitution .

5.

${\mathit{x}}^{\mathbf{2}}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}{\left(y^{\text{'}}\right)}^{\mathbf{2}}\mathbf{=}\mathbf{0}$

Short answer

The solution is

$\frac{1}{c_1}x+\frac{1}{c_1^2}\ln \left|c_{1}x-1\right|+C$

Step-by-step solution

given information

$x^2{y}^{\text{'}\text{'}}+{\left(y^{\text{'}}\right)}^2=0,$

solution

$$\begin{gathered} x^2{y}^{\text{'}\text{'}}+{\left(y^{\text{'}}\right)}^2=0,y^{\text{'}}=u,y^{\text{'}\text{'}}=u^{\text{'}} \\ {x}^2{u}^{\text{'}}+u^2=0 \\ {x}^2{u}^{\text{'}}=-u^2 \\ \frac{du}{dx}=-x^{-2}{u}^2 \\ -\int u^{-2}du=\int x^{-2}dx \\ {u}^{-1}=-x^{-1}+c_1 \\ u=\frac{1}{c_1-x^{-1}} \\ {y}^{\text{'}}=u\to y=\int \frac{1}{c_1-x^{-1}}dx \\ y=\int \frac{1}{c_1-x^{-1}}dx \\ =\int \frac{x}{c_{1}x-1}dx \\ {c}_{1}x-1=t \\ \Rightarrow dx=\frac{dt}{c_1} \end{gathered}$$

Therefore, integral becomes

$$\begin{gathered} =\int \frac{\frac{1+t}{c_1}}{t}\frac{dt}{c_1} \\ =\frac{1}{c_1^2}\int \left(\frac{1}{t}+1\right)dt \\ =\frac{1}{c_1^2}\left(ln\left(t\right)+t\right)+c \\ =\frac{1}{c_1^2}\left(ln\left(c_{1}x-1\right)+c_{1}x-1\right)+c \\ =\frac{1}{c_1}x+\frac{1}{c_1^2}ln\left(c_{1}x-1\right)+C \end{gathered}$$

conclusion

The solution is

$\frac{1}{c_1}x+\frac{1}{c_1^2}\ln \left|c_{1}x-1\right|+C$