Answer:
[tex]y(x)=x^2+5x[/tex]
Step-by-step explanation:
Given: [tex]y'=\sqrt{4y+25}[/tex]
Initial value: y(1)=6
Let [tex]y'=\dfrac{dy}{dx}[/tex]
[tex]\dfrac{dy}{dx}=\sqrt{4y+25}[/tex]
Variable separable
[tex]\dfrac{dy}{\sqrt{4y+25}}=dx[/tex]
Integrate both sides
[tex]\int \dfrac{dy}{\sqrt{4y+25}}=\int dx[/tex]
[tex]\sqrt{4y+25}=2x+C[/tex]
Initial condition, y(1)=6
[tex]\sqrt{4\cdot 6+25}=2\cdot 1+C[/tex]
[tex]C=5[/tex]
Put C into equation
Solution:
[tex]\sqrt{4y+25}=2x+5[/tex]
or
[tex]4y+25=(2x+5)^2[/tex]
[tex]y(x)=\dfrac{1}{4}(2x+5)^2-\dfrac{25}{4}[/tex]
[tex]y(x)=x^2+5x[/tex]
Hence, The solution is [tex]y(x)=\dfrac{1}{4}(2x+5)^2-\dfrac{25}{4}[/tex] or [tex]y(x)=x^2+5x[/tex]
