B uchun yechish
\left\{\begin{matrix}\\B=0\text{, }&\text{unconditionally}\\B\in \mathrm{R}\text{, }&T=0\end{matrix}\right,
K uchun yechish
K\in \mathrm{R}
T=0\text{ or }B=0
Baham ko'rish
Klipbordga nusxa olish
1TB-\frac{\mathrm{d}}{\mathrm{d}x}(M)KB=0
Ikkala tarafdan \frac{\mathrm{d}}{\mathrm{d}x}(M)KB ni ayirish.
BT-BK\frac{\mathrm{d}}{\mathrm{d}x}(M)=0
Shartlarni qayta saralash.
-BK\frac{\mathrm{d}}{\mathrm{d}x}(M)+BT=0
Shartlarni qayta saralash.
\left(-K\frac{\mathrm{d}}{\mathrm{d}x}(M)+T\right)B=0
B'ga ega bo'lgan barcha shartlarni birlashtirish.
TB=0
Tenglama standart shaklda.
B=0
0 ni T ga bo'lish.
\frac{\mathrm{d}}{\mathrm{d}x}(M)KB=1TB
Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.
BK\frac{\mathrm{d}}{\mathrm{d}x}(M)=BT
Shartlarni qayta saralash.
0=BT
Tenglama standart shaklda.
K\in
Bu har qanday K uchun xato.
Misollar
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