Solve y'''-y'' +4y'-4y=40(t^2+t+1)H(t-2). y(0)=5, y'(0)=0, y''(0)=10 (Impluse Function)
CHP:-solving differential equation using laplace transform and matrix
Answers
Answer:
Moment of inertia of disc about an axis passing through center and perpendicular to plane is 1/2(M×R^2) where M is mass of disc and R is radius. ( Derivation are available in books for 11th NCERT book and 12th state boards, as well as different CET and competitive exam books)
MI(moment of inertia) about axis passing through tangent and perpendicular to plane is (3/2) MR^2 (using principle of parallel axes)
MI about axis passing through diameter in plane of disc is (1/4) MR^2 (using principle of perpendicular axes)
MI about axis passing through tangent, in plane of disc is (5/4) MR^2 (using principle of parallel axes).Mauryan rulers favored and promoted mainly non-Hindu religions; whereas Gupta rulers followed and promoted Hinduism. Great architectural structures and pillars were built during the Mauryan dynasty; whereas science, literature and astronomy flourished during the Gupta period.Let us consider a quadratic polynomial p x^{2}+q x+r=0px
2
+qx+r=0
We know that if discriminant q^{2}-4 p r=0q
2
−4pr=0 , the roots of the equation are equal.
Here q=-2\left(a^{2}-b c\right)q=−2(a
2
−bc) ; p=c^{2}-a bp=c
2
−ab ; r=b^{2}-a cr=b
2
−ac
Therefore, discriminant=0
\begin{gathered}\begin{array}{l}{q^{2}=4 p r} \\ \\{\left(-2\left(a^{2}-b c\right)\right)^{2}=4\left(c^{2}-a b\right)\left(b^{2}-a c\right)} \\ \\{4\left(a^{4}+b^{2} c^{2}-2 a^{2} b c\right)=4\left(b^{2} c^{2}+a^{2} b c-a b^{3}-a c^{3}\right)} \\ \\{a^{4}+b^{2} c^{2}-2 a^{2} b c=b^{2} c^{2}+a^{2} b c-a b^{3}-a c^{3}}\end{array}\end{gathered}
q
2
=4pr
(−2(a
2
−bc))
2
=4(c
2
−ab)(b
2
−ac)
4(a
4
+b
2
c
2
−2a
2
bc)=4(b
2
c
2
+a
2
bc−ab
3
−ac
3
)
a
4
+b
2
c
2
−2a
2
bc=b
2
c
2
+a
2
bc−ab
3
−ac
3
\begin{gathered}\begin{array}{l}{a^{4}+b^{2} c^{2}-2 a^{2} b c-b^{2} c^{2}-a^{2} b c+a b^{3}+a c^{3}=0} \\ \\{a\left(a^{3}+b^{3}+c^{3}\right)=3 a^{2} b c} \\ \\{a\left(a^{3}+b^{3}+c^{3}-3 a b c\right)=0}\end{array}\end{gathered}
a
4
+b
2
c
2
−2a
2
bc−b
2
c
2
−a
2
bc+ab
3
+ac
3
=0
a(a
3
+b
3
+c
3
)=3a
2
bc
a(a
3
+b
3
+c
3
−3abc)=0
\begin{gathered}\begin{array}{l}{a=0 \text { or } a^{3}+b^{3}+c^{3}-3 a b c=0} \\ \\{a=0 \text { or } a^{3}+b^{3}+c^{3}=3 a b c}\end{array}\end{gathered}
a=0 or a
3
+b
3
+c
3
−3abc=0
a=0 or a
3
+b
3
+c
3
=3abc