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Use nested conditions in the blade

I'm making a horizontal navigation menu in Blade (Laravel 5.7) with the typical effect that the option corresponding to the current page looks different, and in my case the link has been removed, without any problem with a code like this for each menu item:


    
    
    
    @if (Request :: is (& # 39; / & # 39;))
        At home
    @else
        
            At home
    @endif

The doubt has arisen because one of the options of the menu is displayed according to a value set in the configuration or not.
I was able to solve it @unless:

@unless (! Config :: get (& # 39; app.events & # 39;))
    
        @if (Request :: is (& # 39; events & # 39;))
            Events
        @else
            
                Events
        @endif
    
@endunless

but I started testing with:

@if (config :: get (& # 39; app.events & # 39;))
    
        @if (Request :: is (& # 39; events & # 39;))
            Events
        @else
            
                Events
        @endif
    
@endif

This last code cascaTherefore, my question arises.

If you need to use conditional nesting in Bladeas in the previous example Is there a possibility that is not in a block? @php?

@php
// php code ...
@endphp

Second order differential equations with initial conditions solved by Laplace transforms

Below is a problem that I found in the book Differential Equations by K.A. Stroud and Dexter Booth. I've got the right answer, but I'm not sure I did it right, especially if I took the inverse Laplace transform. So I hope someone can check this for me.
Many Thanks,
bob

Problem:
Solve the following differential equation:
$$ f & # 39; & # 39; (x) – 5f ((x) + 6f (x) = e ^ {2x} text {where} f (0) = 0 text {and} f & # 39; (0) = 0 $$
New line
Reply:
To solve this equation, I use the Laplace transform.
begin {align *}
s ^ 2F (s) – sf (0) – f ((0) – 5 (sF (s) – f (0)) + 6F (s) & = frac {1} {s-2} \
s ^ 2F (s) – 5 (sF (s)) + 6F (s) & = frac {1} {s-2} \
(s ^ 2-5s + 6) F (s) & = frac {1} {s-2} \
F (s) & = frac {1} {(s-2) ^ 2 (s-3)} \
end {align *}
Now we apply the technique of fractions to find them $ f (x) $,
begin {align *}
frac {1} {(s-2) ^ 2 (s-3)} & = frac {A} {s-2} + frac {B} {(s-2) ^ 2} + frac { C} {s-3} \
1 & = A (s-2) (s-3) + B (s-3) + C (s-2) ^ 2
end {align *}
At the $ s = 2 $ we have: $ 1 = B (2-3) = -B $ or $ B = -1 $, At the $ s = 3 $ we have L $ 1 = C (3-2) ^ 2 = C = 1 $,
begin {align *}
A (s-2) (s-3) – (s-3) + (s-2) ^ 2 & = 1 \
A + 1 & = 0 \
A & = -1 \
end {align *}
begin {align *}
F (s) & = frac {-1} {s-2} – frac {1} {(s-2) ^ 2} + frac {1} {s-3} \
mathcal {L} ^ {- 1} left ( frac {1} {s-2} right) & = e ^ {- 2x} \
end {align *}
Now we have to find the inverse Laplace transform of:
$$ frac {1} {(s-2) ^ 2} $$
Remember, if $ mathcal {L} (f (t)) = F (s) $ then $ mathcal {L} (e ^ {at} f (t)) = F (s-a) $
In this case, we use $ a = 2 $, $ f (t) = x $ and $ F (s) = frac {1} {s ^ 2} $,
begin {align *}
mathcal {L} ^ {- 1} left ( frac {1} {(s-2) ^ 2} right) & = xe ^ {2x} \
mathcal {L} ^ {- 1} left ( frac {1} {s-3} right) & = e ^ {3x} \
f (x) & = -e ^ {- 2x} – xe ^ {- 3x} + e ^ {3x} \
f (x) & = e ^ {3x} – (x + 1) e ^ {2x} \
end {align *}