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Alex17521 [72]
3 years ago
12

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

Mathematics
1 answer:
goblinko [34]3 years ago
4 0

Answer:

y(x)=x^2+5x

Step-by-step explanation:

Given: y'=\sqrt{4y+25}

Initial value: y(1)=6

Let y'=\dfrac{dy}{dx}

\dfrac{dy}{dx}=\sqrt{4y+25}

Variable separable

\dfrac{dy}{\sqrt{4y+25}}=dx

Integrate both sides

\int \dfrac{dy}{\sqrt{4y+25}}=\int dx

\sqrt{4y+25}=2x+C

Initial condition, y(1)=6

\sqrt{4\cdot 6+25}=2\cdot 1+C

C=5

Put C into equation

Solution:

\sqrt{4y+25}=2x+5

or

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

y(x)=\dfrac{1}{4}(2x+5)^2-\dfrac{25}{4}

y(x)=x^2+5x

Hence, The solution is y(x)=\dfrac{1}{4}(2x+5)^2-\dfrac{25}{4} or y(x)=x^2+5x

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