Engineering MathematicsΒ is one of the scoring sections inΒ GATE/BARC/ISRO Exam. Looking at your requirement, we are sharing with you Important Engineering Mathematics Formulas & Shortcuts for Competitive Exam as well as Engineering exam.
- IfΒ πβ πβ², no solution, ifΒ π=πβ²=π, unique solution ifΒ π=πβ²<π, many solutions. (non-homogeneous)
- IfΒ π=π, trivial solution, ifΒ π<πΒ ,then (πβπ) linearly independent solutions. (Many solutions) and ifΒ π<π, then many solutions.
- π(π₯+β) =π(π₯) +βπβ²(π₯) +β22!πβ²β²(π₯) +β33!πβ²β²β²(π₯) + β¦β¦β¦β
- IfΒ ππ‘βπ 2>0 andΒ π<0Β π(π₯,π¦) have maximum, ifΒ ππ‘βπ 2>0 andΒ π>0Β π(π₯,π¦) have minimum at(π,π) and ifΒ ππ‘βπ 2<0, then saddle point. IfΒ ππ‘βπ 2=0,ππ’ππ‘βππΒ investigation is required to decide.
- β«π(β ππ₯+πππ¦)=β«β«πΈ(ππππ₯βπβ ππ¦)ππ₯ππ¦Β (Greenβs)
- β«ππ.πβ=β«πππ’πππ.πππ Β (Stokes)
- β«ππ.πππ =β«πΈπππ£πππ£Β (Gauss)
- π¦πβ«πππ₯=β«ππβ«πππ₯ππ₯+π
- IfΒ πΒ ππ₯+πΒ ππ¦=0 be a homogeneous equation inΒ π₯Β andΒ π¦, then 1πΒ π₯+πΒ π¦Β ππ Β an integrating factor
- If the equation of the typeΒ π1(π₯π¦)π¦Β ππ₯+π2(π₯π¦)π₯Β ππ¦=0. If the equationΒ πΒ ππ₯+πΒ ππ¦=0 be of this type then 1πΒ π₯βπΒ π¦Β is an integrating factor
- IfΒ ππππ¦βππππ₯πΒ be a function of x only =Β π(π₯) say thenΒ πβ«π(π₯)ππ₯Β is an integrating factor
- IfΒ ππππ₯βππππ¦πΒ be a function of y only =Β π(π¦) say thenΒ πβ«π(π¦)ππ¦Β is an integrating factor.
- β«πππ₯π¦=ππππ‘Β +β«terms of N not containing x dy=c
- π.πΌ.=1π(π·)πππ₯=1π(π)πππ₯Β ,Β π(π)β 0, ifΒ π(π)=0,π‘βππΒ π.πΌ.=π₯1πβ²(π)πππ₯,πβ²(π)β 0
- π.πΌ=1π(π·2)sin(ππ₯+π)= 1π(βπ2),Β π(βπ2)β 0, ifΒ π(βπ2)=0,
thenΒ π.πΌ. =π₯1πβ²(βπ2)sin(ππ₯+π),Β πβ²(βπ2)β 0
- π.πΌ.=1π(π·)πππ₯π=Β πππ₯1π(π·+π)π
- π.πΌ=1π(π·)π₯π=[π(π·)]β1π₯π,
- (1+π₯)β1=1βπ₯+π₯2ββ―
- (1βπ₯)β1=1+π₯+π₯2+β―
- π₯ππππ¦ππ₯π+π1π₯πβ1ππβ1π¦ππ₯πβ1+β―ππβ1π₯ππ¦ππ₯+πππ¦=π,Β π₯=ππ‘Β ,Β π₯ππ¦ππ₯=π·π¦,Β π₯2π2π¦ππ₯2=π·(π·β1)π¦,Β π₯3π3π¦ππ₯3=D(Dβ1)(Dβ2)
- πππ₯(β«β(π‘,π₯)ππ‘π(π₯)π(π₯))=β«πππ₯β(π‘,π₯)ππ‘π(π₯)π(π₯)+ππππ₯Β β[π(π₯),π₯]βππππ₯Β β[π(π₯),π₯]
- πΏ{π(π‘)}=β«πβπ π‘β0Β π(π‘)ππ‘
- πΏ(1)=1π
- πΏ(π‘π)=π!π π+1
- πΏ(πππ‘)=1π βπ
- πΏ(sinππ‘)=ππ 2+π2
- πΏ(cosππ‘)=π π 2+π2
- πΏ(sinhππ‘)=ππ 2βπ2
- πΏ(coshππ‘)=π π 2βπ2
- πΏ{πππ‘π(π‘)}=Β πΜ (π βπ)
- π(π‘+π)=π(π‘) thenΒ πΏ{π(π‘)}= β«πβπ π‘Β π(π‘)ππ‘π01βπβπ π
- πΏ{πβ²(π‘)}=π πΜ (π )βπ(0)
- πΏ{ππ(π‘)}=Β π ππΜ (π )βπ πβ1π(0)βπ πβ2πβ²(0)ββ―β¦β¦..ππβ1(0)
- πΏ{β«π(π₯)ππ₯π‘0}= 1ππΜ (π )
- πΏ{π‘ππ(π‘)}=(β1)πππππ π.[πΜ (s)]
- πΏ{1π‘π(π‘)}= β«πΜ (s) βπππ
- π(π₯)=π02+ Ξ£ππβπ=1cosππ₯+ Ξ£ππβπ=1sinππ₯
- π0=1πβ«π(π₯)ππ₯πΌ+2ππΌ,Β ππ=1πβ«π(π₯) cosππ₯ππ₯πΌ+2ππΌ,Β ππ=1πβ«π(π₯) sinππ₯ππ₯πΌ+2ππΌ
- π(π₯)=π02+ Ξ£ππβπ=1cosπππ₯π+ Ξ£ππβπ=1sinπππ₯π
- 40.Β π0=1πβ«π(π₯)ππ₯πΌ+2ππΌ,Β ππ=1πβ«π(π₯) cosπππ₯πππ₯πΌ+2ππΌ,Β ππ=1πβ«π(π₯) sinπππ₯πππ₯πΌ+2ππΌ
- π(π₯)= Ξ£ππβπ=1sinπππ₯πΒ , whereΒ ππ=2πβ«π(π₯) sinπππ₯πππ₯π0
- π(π₯)=Β π02+ Ξ£ππβπ=1cosπππ₯πΒ where,Β π0=2πβ«π(π₯)ππ₯π0,Β ππ=2πβ«π(π₯) cosπππ₯πππ₯π0
- π=Σ π₯ππ(π₯π)πΒ πππΒ π=β«π₯Β π(π₯)ππ₯Β βββ
- π2=Ξ£ (π₯πβπ)2π(π₯π)Β ππππΒ π2=β«(π₯βπ)2Β π(π₯)ππ₯βββ
- ππππ:ππ=πΒ ππππππππ:Β π2=πΒ (Poissonβs distribution)
- π(π₯)=1πβ2ππβ 12(π₯βππ)2 (Normal distribution)
- π¦=π+ππ₯, Ξ£π¦=ππ+πΞ£π₯, Ξ£π₯π¦=πΞ£π₯+πΞ£π₯2
- π¦=π+ππ₯+ππ₯2, Ξ£π¦=ππ+π.Ξ£π₯+π.Ξ£π₯2, Ξ£π₯π¦=πΞ£π₯+π.Ξ£π₯2+πΒ Ξ£π₯3, Ξ£π₯2π¦=πΞ£π₯2+π.Ξ£π₯3+πΒ Ξ£π₯4
- π¦=π(π₯)=(π₯βπ₯1)(π₯βπ₯2)β¦(π₯βπ₯π)(π₯0βπ₯1)(π₯0βπ₯2)β¦(π₯0βπ₯π)π¦0+(π₯βπ₯0)(π₯βπ₯2)β¦(π₯βπ₯π)(π₯1βπ₯0)(π₯1βπ₯2)β¦(π₯1βπ₯π)π¦1+β―+(π₯βπ₯1)(π₯βπ₯2)β¦(π₯βπ₯πβ1)(π₯πβπ₯0)(π₯πβπ₯1)β¦(π₯πβπ₯πβ1)π¦π
- (ππ¦ππ₯)π₯0=1β[Ξπ¦0β12Ξ2π¦0+13Ξ3π¦0β14Ξ4π¦0+β―]
- (ππ¦ππ₯)π₯π=1β[βπ¦π+12β2π¦π+13β3π¦π+14β4π¦π+β―]
- π₯π+1=π₯πβπ(π₯π)πβ²(π₯π) (Newton-Raphson)
- β«π(π₯)ππ₯π₯0+πβπ₯0=β2[π¦0+π¦π+2(π¦1+π¦2+β―..+π¦πβ1)] (Trapezoidal)
- β«π(π₯)ππ₯π₯0+πβπ₯0=β3[(π¦0+π¦π)+4(π¦1+π¦3+β―π¦πβ1)+2(π¦2+π¦4+β―π¦πβ2)] (Simpsonβs)
- πΈππππ=βπβπ12β2Β πβ²β²(π)=π(β2) (Trapezoidal)
- πΈππππ=βπβπ180β4πππ£(π)=π(β4) (Simpsonβs)
- π¦π+1=π¦π+β.π(π‘π,π¦π) whereΒ ππ¦ππ₯=π(π‘,π¦) (Eulerβs)
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