Solve the given initial-value problem for y0 > 0. dy dx = y , y(x0) = y0 y(x) = find the largest interval i on which the solution is defined. (enter your answer using interval notation.)

Respuesta :

The largest interval I on which the solution is defined.

[tex]y(x)=\left(\frac{x}{2}+\sqrt{y_{0}}-\frac{x_{0}}{2}\right)^{2}[/tex]

This is further explained below.

Find the largest interval I on which the solution is defined. ?

Generally, the equation for is  mathematically given as

[tex]\frac{d y}{d x}=\sqrt{y}[/tex]

Separating variables and integrating

[tex]\begin{aligned}&\Rightarrow \int \frac{d y}{\sqrt{y}}=\int d x \\&\Rightarrow 2 \sqrt{y}=x+c \\&\text { Put } y\left(x_{0}\right)=y_{0} \\&\Rightarrow 2 \sqrt{y_{0}}=x_{0}+c \\&\Rightarrow c=2 \sqrt{y_{0}}-x_{0}(2) \\&\text { Put }(2) \text { in } \\&\Rightarrow 2 \sqrt{y}=x+2 \sqrt{y_{0}}-x_{0} \\&\Rightarrow \sqrt{y}=\frac{x}{2}+\sqrt{y_{0}}-\frac{x_{0}}{2} \\&\Rightarrow y(x)=\left(\frac{x}{2}+\sqrt{y_{0}}-\frac{x_{0}}{2}\right)^{2} \end{aligned}[/tex]

In conclusion, the largest interval I on which the solution is defined.

[tex]y(x)=\left(\frac{x}{2}+\sqrt{y_{0}}-\frac{x_{0}}{2}\right)^{2}[/tex]

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