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 u = y'.

4.

${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{=}\mathbf{1}\mathbf{+}{\left(y^{\text{'}}\right)}^{\mathbf{2}}$

Short answer

The solution is

$y=-ln\left(cos\right(x+c\left)\right)+k$

Step-by-step solution

given information

$y^{\text{'}\text{'}}=1+{\left(y^{\text{'}}\right)}^2$

solve the equation

We have the second order differential equation

$y^{\text{'}\text{'}}=1+{\left(y^{\text{'}}\right)}^2.....\left(1\right)$

and we have to solve it as the following technique:

First, we have the substitution

$y^{\text{'}}=u.....\left(2\right)$

After that, since we have y' = u and y is a dependent variable then by differentiation with respect to x we can have

$y^{\text{'}\text{'}}=u^{\text{'}}.....\left(3\right)$

Second, substitute from equations (2) (y') and (y'') (3) with into the given equation (1) , then we obtain

$$\begin{gathered} u^{\text{'}}=1+(u{)}^2 \\ \frac{du}{dx}=1+u^2 \end{gathered}$$

After that, we have to separate variables and do integration as

$$\begin{gathered} \int \frac{1}{1+u^2}du=\int dx \\ x+c=\int \frac{1}{1+u^2}du.....\left(4\right) \end{gathered}$$

substitution

This integration $\int \frac{1}{1+u^2}$can be made by substitution as the following:

If we set $u=tanv=\frac{sinv}{cosv}$, then we have

$$\begin{gathered} \int \frac{1}{1+u^2}du=\int \frac{1}{1+\frac{si{n}^{2}v}{co{s}^{2}v}}\frac{1}{co{s}^{2}v}du \\ =\int \frac{1}{si{n}^{2}v+co{s}^{2}v}\frac{1}{co{s}^{2}v}du \\ =\int \frac{co{s}^{2}v}{si{n}^{2}v+co{s}^{2}v}\frac{1}{co{s}^{2}v}du \\ =\int \frac{co{s}^{2}v}{1}\frac{1}{co{s}^{2}v}du \\ =\int \frac{co{s}^{2}v}{co{s}^{2}v}du \\ =\int dv \\ =v+c \\ =ta{n}^{-1}\left(u\right) \end{gathered}$$

Now, substitute with the result of this integration into equation (4), then we obtain

$x+c=ta{n}^{-1}\left(u\right)$

Then we have

$u=tan(x+c).....\left(5\right)$

use integration

Third, substitute from equation (2) with the result of into equation (5) , then we have

$$\begin{gathered} y^{\text{'}}=tan(x+c) \\ \frac{dy}{dx}=tan(x+c) \end{gathered}$$

After that, we have to separate variables and do integration as

$$\begin{gathered} dy=tan(x+c)dx \\ \int dy=\int tan(x+c)dx \\ \int dy=\int \frac{sin(x+c)}{cos(x+c)} \\ \int dy=-\int \frac{-sin(x+c)}{cos(x+c)} \\ y=-ln\left(cos\right(x+c\left)\right)+k \end{gathered}$$

conclusion

The solution is

$y=-ln\left(cos\right(x+c\left)\right)+k$