Question

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 1

`y(x)=x^2+5x`

Explanation:

Given:
`y'=√(4y+25)`

Initial value: y(1)=6

Let
`y'=(dy)/(dx)`

`(dy)/(dx)=√(4y+25)`

Variable separable

`(dy)/(√(4y+25))=dx`

Integrate both sides

`\int (dy)/(√(4y+25))=\int dx`

`√(4y+25)=2x+C`

Initial condition, y(1)=6

`√(4\cdot 6+25)=2\cdot 1+C`

`C=5`

Put C into equation

Solution:

`√(4y+25)=2x+5`

or

`4y+25=(2x+5)^2`

`y(x)=(1)/(4)(2x+5)^2-(25)/(4)`

`y(x)=x^2+5x`

Hence, The solution is
`y(x)=(1)/(4)(2x+5)^2-(25)/(4)` or
`y(x)=x^2+5x`