Answer:
Hence the domain of the function y = 5 tan 5x is ,
[tex]\large \boxed {R-\{(2n+1)\frac{\pi}{10},n-1,2,...\} }[/tex]
and largest interval of definition for the solution is [tex]\large \boxed {-\frac{\pi}{10},\frac{\pi}{10} }[/tex]
Step-by-step explanation:
Considering the differential equation
[tex]y' = 25 + y^2---(1)[/tex]
and the function y = 5 tan 5x
Find the domain of the function y = 5 tan 5x
Since [tex]y=\frac{5 \sin 5x}{\cos 5x}[/tex] then determine the zeros of the denominator, cos 5x
The zeros of the denominator of cos 5x is [tex]\{ x:x=(2n+1)\frac{\pi}{10} \}[/tex]
Hence the function y = 5 tan 5x exist on [tex]R-\{(2n+1)\frac{\pi}{10} ,n=1,2...\}[/tex]
Therefore, the domain of the function y = 5 tan 5x is [tex]R-\{(2n+1)\frac{\pi}{10} ,n=1,2...\}[/tex]
Differential the function y = 5 tan 5x with respect to x
[tex]\frac{d}{dx} y=\frac{d}{dx} (5 \tan 5x)\\\\y'=5\sec^25x\frac{d}{dx} (5x)\\\\ \text {since }\frac{d}{du} (\tan u)= \sec^2 u\\\\y'=25 \sec^25x[/tex]
Use trigonometry identity , [tex]1+\tan^2 \theta =\sec^2 \theta[/tex]
Substitute [tex]1+\tan ^25x\ \text{for} \sec^25x[/tex]
[tex]y'=25(1+\tan^25x)\\y'=25+25\tan^25x\\y'=25+(5\tan5x)^2\\y'=25+y^2[/tex]
Therefore the function y = 5 tan 5x satisfies the differential equation [tex]y'=25+y^2[/tex]
Now , Find the largest interval of the solution
Since the tangent function is periodic with period π so take interval
[tex]-\frac{\pi}{2} <5x<\frac{\pi}{2} \\\\-\frac{\pi}{10} <x<\frac{\pi}{10}[/tex]
since , the function y = 5 tan 5x is not differentiatable at [tex]x = -\frac{\pi}{10}\ \text {and} \ x=\frac{\pi}{10}[/tex]
Hence the domain of the function y = 5 tan 5x is ,
[tex]\large \boxed {R-\{(2n+1)\frac{\pi}{10},n-1,2,...\} }[/tex]
and largest interval of definition for the solution is [tex]\large \boxed {-\frac{\pi}{10},\frac{\pi}{10} }[/tex]
