Question

Solve the initial value problem.?
dy/dx = x*sqrt(x-4) and y=5 when x=4. I have an answer key, but I cannot get my answer to match!

Answer 1

@SpeirsTheAmazingHD solving a moment pls !

Answer 2

\[dy/dx=x \sqrt{x-4}\]
\[dy=x \sqrt{x-4} dx\]
\[y=\int\limits(x \sqrt{x-4})dx\]
\[Let, u =\sqrt{x-4} \]
\[then , x=u^2+4\]
\[and, \frac{ du }{ dx }= \frac{ 1 }{ 2\sqrt{x-4} }\]
\[and , dx=(2\sqrt{x-4})du\]
\[y=\int\limits (x \sqrt{x-4} )dx \]
\[y=\int\limits ((u^2+4)(\sqrt{x-4})(2\sqrt{x-4}))du\]
\[y=2\int\limits (u^2+4)(x-4) du\]
\[y=2\int\limits (u^2+4)(u^2+4-4) du\]
\[y=2\int\limits (u^2+4)(u^2)du\]
\[y=\int\limits (2u^4+8u^2 )du\]
\[y=\frac{ 2 }{ 5 }u^5 + \frac{ 8 }{ 3 }u^3 + c\]
\[y=\frac{ 2 }{ 5 }(x-4)^\frac{ 5 }{ 2 } + \frac{ 8 }{ 3 }(x-4)^\frac{ 3 }{ 2 } + c\]

Answer 3

Put value of y=5 and x=4 and get value of c then put c into the equation this will be the solution to the initial value problem

Answer 4

You will get c=5
Hope you understood :)