求解$ \ dfrac {dy} {dx} = \ cos(x + y)+ \ sin(x + y)$

ss1729 06/18/2018. 2 answers, 86 views
integration differential-equations trigonometry indefinite-integrals

求解$ \ frac {dy} {dx} = \ cos(x + y)+ \ sin(x + y)$

My Attempt $$ \ frac {dy} {dx} = \ sqrt {2} \ cos(x + y- \ tfrac {\ pi} {4})$$设置$ t = x + y- \ tfrac {\ pi } {4} \暗示y = t-x + \ tfrac {\ pi} {4} $ $$ \ frac {dy} {dx} = \ frac {dt} {dx} -1 = \ sqrt {2} \ cos t \\ \ frac {dt} {dx} = \ sqrt {2} \ cos(t)+1 \\ \ int \ frac {dt} {\ sqrt {2} \ cos(t)+1} = \ int dx $$如何进一步查找常规解决方案?

2 Answers


Cesareo 06/18/2018.

将$ x + y = u \设为{frac {dy} {dx} = \ frac {du} {dx} -1 $

$$ \ frac {du} {dx} = \ sin u + \ cos u +1 $$

这是一个可分离的DE

$$ \ frac {du} {\ sin u + \ cos u + 1} = dx $$

等等

注意

$$ \ sin u + \ cos u = \ sqrt 2 \ sin(u + \ frac {\ pi} {4})$$

现在使用身份

$$ \ cos u = 2 \ cos ^ 2(\ frac u2)-1 \\ \ sin u = 2 \ cos(\ frac u2)\ sin(\ frac u2)$$

我们有

$$ \ frac {1} {\ cos u + \ sin u + 1} = \ frac {1} {2 \ cos(\ frac u2)(\ cos(\ frac u2)+ \ sin(\ frac u2))} = \ frac 12 \ left(\ frac {\ sin(\ frac u2)} {\ cos(\ frac u2)} + \ frac { - \ sin(\ frac u2)+ \ cos(\ frac u2)} {\ cos(\ frac u2)+ \ sin(\ frac u2)} \ right)$$

等等


ss1729 06/19/2018.

$$ \ begin {align}&\ int dx = \ int \ frac {dt} {\ sqrt {2} \ cos(t)+1} \\ x&= \ frac {1} {\ sqrt {2}} \ int \ frac {dt} {\ cos t + \ frac {1} {\ sqrt {2}}} = \ frac {1} {\ sqrt {2}} \ int \ frac {dt} {\ cos t + \ cos \压裂{\ PI} {4}} = \压裂{1} {\ SQRT {2}} \ INT \压裂{DT} {2个\ COS \大(\压裂{吨} {2} + \压裂{\ PI } {8} \ big)\ cos \ big(\ frac {t} {2} - \ frac {\ pi} {8} \ big)} \\&= \ frac {\ sqrt {2}} {2 \ SQRT {2}} \ INT \压裂{\罪\大[\大(\压裂{吨} {2} + \压裂{\ PI} {8} \大) - \大(\压裂{吨} {2 } - \压裂{\ PI} {8} \大)\大]} {\ COS \大(\压裂{吨} {2} + \压裂{\ PI} {8} \大)\ COS \大( \ frac {t} {2} - \ frac {\ pi} {8} \ big)} dt \\&= \ frac {1} {2} \ bigg [\ int \ tan \ Big(\ frac {t} {2} + \压裂{\ PI} {8} \大)DT- \ INT \黄褐色\大(\压裂{吨} {2} - \压裂{\ PI} {8} \大)DT \比格] \\&\ text {Set} u = \ frac {t} {2} + \ frac {\ pi} {8} \暗示dt = 2du \ quad \&\ quad v = \ frac {t} {2} - \ frac {\ pi} {8} \暗示dt = 2dv \\ x&= \ int \ tan u.du- \ int \ tan v.dv = \ log | \ sec u | - \ log | \ sec v | = \ log | \ frac {\ sec u} {\ sec v} | + C \\&= \ log | \ frac {\ cos v} {\ cos u} | + C = \ log \ bigg | \ frac {\ COS \大(\ tfrac {吨} {2} - \ tfrac {\ PI} {8} \大)} {\ COS \大(\ tfrac {吨} {2} + \ tfrac {\ PI} {8} \ big)} \ bigg | + C \\&= \ log \ bigg | \ frac {\ cos \ big(\ tff 外消旋{X + Y} {2} - \ tfrac {\ PI} {8} - \ tfrac {\ PI} {8} \大)} {\ COS \大(\ tfrac {X + Y} {2} - \ tfrac {\ PI} {8} + \ tfrac {\ PI} {8} \大)} \比格| + C = \ LOG \比格| \压裂{\ COS \大(\ tfrac {X + Y} { 2} - \ tfrac {\ pi} {4} \ big)} {\ cos \ big(\ tfrac {x + y} {2} \ big)} \ bigg | + C \\&= \ log \ bigg | \压裂{1} {\ SQRT {2}} + \压裂{1} {\ SQRT {2}} \黄褐色\大(\ tfrac {X + Y} {2} \大)\比格| + C \\ &\ color {red} {x = \ log \ bigg | 1+ \ tan \ big(\ tfrac {x + y} {2} \ big)\ bigg | + K} \ end {align} $$

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