Question

In Problems 39-44 solve the given differential equation subject to the indicated conditions.

${\mathit{y}}^{\mathbf{\text{'}}}{\mathit{y}}^{\mathbf{\text{'}\text{'}}}\mathbf{=}\mathbf{4}\mathit{x}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{5}\mathbf{,}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{2}$

Short answer

The solution is$y=x^2+4$

Step-by-step solution

Given Data

Differential equation is the equation which relate one or more unknown functions and their derivatives.

Given differential equation,

${\mathrm{y}}^{\text{'}}{\mathrm{y}}^{\text{'}\text{'}}=4\mathrm{x},\mathrm{y}\left(1\right)=5,{\mathrm{y}}^{\text{'}}\left(1\right)=2$

Separate variables and integration

We have the non-linear differential equation

${\mathrm{y}}^{\text{'}}{\mathrm{y}}^{\text{'}\text{'}}=4\mathrm{x}$

with the initial conditions

$y\left(1\right)=5\text{and}{y}^{\text{'}}\left(1\right)=2$

and we have to solve it as the following technique:
First, we have to let that

$y^{\text{'}}=u$

After that, since we have$y^{\text{'}}=u$, then by differentiation with respect to$x$, we can have

$y\text{'}\text{'}=u\text{'}$

Second, substitute from equations$\left(2\right)$and$\left(3\right)$with$\left(y\text{'}\right)$ and$\left(y\text{'}\text{'}\right)$ into the given equation, then we obtain

$\begin{array}{rr}& u{u}^{\text{'}}=4x\\ & u\frac{du}{dx}=4x\end{array}$

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

Substitute the values

Third, substitute from equation $\left(2\right)$ with $u$ into equation $\left(4\right)$, then we have

role="math" localid="1667898591759" $y^{\text{'}}=\sqrt{(4x^2+k)}$

Now, we have to apply to apply the point $(x,y^{\text{'}})=(1,2)$ into equation (4) to obtain the constant$\left(k\right)$as

role="math" localid="1667898977293" $\begin{array}{c}2=\sqrt{4+\mathrm{k}}\\ 4=4+\mathrm{k}\\ \mathrm{k}=0\end{array}$

After that, substitute with the value of constant $\left(k\right)$ into equation $\left(4\right)$, then we obtain

$\begin{array}{cc}{y}^{\text{'}}& =\sqrt{4x^2}\\ & =2x\end{array}$

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

$\begin{array}{c}\frac{dy}{dx}=2x\\ dy=2xdx\\ {\int }^{\text{}}dy={\int }^{\text{}}2xdx\end{array}$

Then we have,

$\begin{array}{cc}y& =2\times \frac{1}{2}{x}^2+h\\ & =x^2+h\end{array}$

Final Answer

Now, we have to apply to apply the point$(x,y)=(1,5)$into equation$\left(5\right)$to obtain the constant$\left(h\right)$as

$\begin{array}{c}5={\left(1\right)}^2+h\\ h=5-1\\ h=4\end{array}$

After that, substitute with the value of constant $\left(h\right)$into equation $\left(5\right)$, then we obtain

$y=x^2+4$

is the required solution for the given differential equation at the given conditions.

Thus, the solution is
$y=x^2+4$