Solve the initial value problem: y'(x)=(4y(x)+25)^(1/2) ,y(1)=6. you can't really tell, but the '1/2' is the exponent

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]