Second-Order Linear Non-Homogeneous Equation with constant coefficient

[Mr Adam Gella]

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[Odili]

Maziku Masunga Jr.,

Mkuu salute! Unakata nyanga hatari

 

[Makanyaga]

Msaada:will someone kindly solve this(I am bit mathematically rusty off late)
Ay"+By'+Cy=D+E(sinwt)

Hii site hapa
integral of e^xlnx - Integral Calculator - Symbolab
ina uwezo wa ku-solve both differential and integral calculus ila kwa baadhi ya problems tu. Make sure that you correctly type your problems. See the attached pdf file

 

[Makanyaga]

Msaada:will someone kindly solve this(I am bit mathematically rusty off late)
Ay"+By'+Cy=D+E(sinwt)

FFile limegoma ku-load, lakini unaweza kuuona huo mfano uliopo kwenye site moja kwa moja wa integral ya e^xlnxdx. You either type a diffrerential equation or an integration problem

 

[Charles Mandela]

Maziku Masunga Jr.,

Mkuu salute! Unakata nyanga hatari

Muda haurudi nyuma mkuu!

 

[black hawk87]

Darius na simplemind ishu iko hivi:

Ay" + By' + Cy = D + E(sinwt).

This is a form of equation where A, B, C, D, E and w are constants and t is a parameter.

Now separate the equation:
Ay" + By' + Cy = D........Equation i

Ay" + By' + Cy = E(sinwt).........Equation ii

Then;

For Equation i
Ay" + By' + Cy = D is in form of f(X) = M ,where M is a constant and f(X) = y. Thus y = M

So if y = M; Then y' = 0 and y" = 0. Since we differentiating constant number, all derivatives become zero ( 0 ).

Then substitute the values above in Equation i

From: Ay'' + By' + Cy = D where y = M; y' = 0 and y" = 0

Then: A(0) + B(0) + C(M) = D

Then: C(M) = D , thus the value of M = D/C but M = f(x) = y

Thus the value of y = D/C


For Equation ii


Ay'' + By' + Cy = E(sinwt) is in form of f(X) = M(sinwt) + N(coswt) and f(X) = y.

Thus y = M(sinwt) + N(coswt)

So if, y = M(sinwt) + N(coswt)

Then y' =Mw(coswt) - Nw(sinwt)

Then y'' = -Mw(sinwt) - Nw(coswt)

Substitute the values above into Equation ii

A(-Mw(sinwt) - Nw(coswt)) + B(Mw(coswt) - Nw(sinwt)) + C(M(sinwt) + N(coswt)) = E(sinwt)
-AMw(sinwt) - ANw(coswt) + BMw(coswt) - BNw(sinwt) + CM(sinwt) + CN(coswt) = E(sinwt) + 0(coswt)

By comparing the values of Sine and Cosine:

For Sine

-AMw(sinwt) - BNw(sinwt) + CM(sinwt) = E(sinwt)...... Equation iii

We flash out (sinwt) and the equation will be: -AMw - BNw + CM = E

Then: M(C - Aw) - NBw = E ;

But (C - Aw) and Bw are both constants.

Then let (C - Aw) be m and Bw be m. Since all values are both constant.

Thus Mm - Nm = E....... Equation iv


For Cosine

- ANw(coswt) + BMw(coswt) + CN(coswt) = 0(coswt)...... Equation v

We flash out (coswt) and the equation will be:

-ANw + BMw + CN = 0

N(C - Aw) + BMw = 0

But (C - Aw) & Bw are both constants.

Then Let (C - Aw) be n and Bw be n.

Thus: Nn + Mn = 0 ..... Equation vi

Thus by combining Equation iv & Equation vi we get;
Mm - Nm = E....... Equation iv
Nn + Mn = 0 ........ Equation vi

Now, let us solve simultaneous equation to obtain the values of M & N where m and n are both constants.

In Equation vi : Nn = -Mn

Thus N = -M, Since n and n are constants. And any valuable multiplied by constant, the answer will be constant.

So let constant M(n) be k .

Therefore the value of N will be - k.

Substitute the value of N = -k into Equation iv
Mm - Nm = E;

We get Mm + km = E
Then Mm = E - km
Then M = (E - km)/m

So let constant (E - km)/m be k .

Therefore the value of M will be k.

Thus the value of y = k(sinwt) - k(coswt)

Since the value of y = y1 + y2. The overall value of y will be given as

y = D/C + k2(sinwt) - k1(coswt) where C, D, k1 and k2 are both cosntants

Muda wa kusolve hii mambo nishaingiza sh ngapi dukan kwa mama godiii

kilicho akilini kitumie

 

[dronedrake]

Darius na simplemind ishu iko hivi:

Ay" + By' + Cy = D + E(sinwt).

This is a form of equation where A, B, C, D, E and w are constants and t is a parameter.

Now separate the equation:
Ay" + By' + Cy = D........Equation i

Ay" + By' + Cy = E(sinwt).........Equation ii

Then;

For Equation i
Ay" + By' + Cy = D is in form of f(X) = M ,where M is a constant and f(X) = y. Thus y = M

So if y = M; Then y' = 0 and y" = 0. Since we differentiating constant number, all derivatives become zero ( 0 ).

Then substitute the values above in Equation i

From: Ay'' + By' + Cy = D where y = M; y' = 0 and y" = 0

Then: A(0) + B(0) + C(M) = D

Then: C(M) = D , thus the value of M = D/C but M = f(x) = y

Thus the value of y = D/C


For Equation ii


Ay'' + By' + Cy = E(sinwt) is in form of f(X) = M(sinwt) + N(coswt) and f(X) = y.

Thus y = M(sinwt) + N(coswt)

So if, y = M(sinwt) + N(coswt)

Then y' =Mw(coswt) - Nw(sinwt)

Then y'' = -Mw(sinwt) - Nw(coswt)

Substitute the values above into Equation ii

A(-Mw(sinwt) - Nw(coswt)) + B(Mw(coswt) - Nw(sinwt)) + C(M(sinwt) + N(coswt)) = E(sinwt)
-AMw(sinwt) - ANw(coswt) + BMw(coswt) - BNw(sinwt) + CM(sinwt) + CN(coswt) = E(sinwt) + 0(coswt)

By comparing the values of Sine and Cosine:

For Sine

-AMw(sinwt) - BNw(sinwt) + CM(sinwt) = E(sinwt)...... Equation iii

We flash out (sinwt) and the equation will be: -AMw - BNw + CM = E

Then: M(C - Aw) - NBw = E ;

But (C - Aw) and Bw are both constants.

Then let (C - Aw) be m and Bw be m. Since all values are both constant.

Thus Mm - Nm = E....... Equation iv


For Cosine

- ANw(coswt) + BMw(coswt) + CN(coswt) = 0(coswt)...... Equation v

We flash out (coswt) and the equation will be:

-ANw + BMw + CN = 0

N(C - Aw) + BMw = 0

But (C - Aw) & Bw are both constants.

Then Let (C - Aw) be n and Bw be n.

Thus: Nn + Mn = 0 ..... Equation vi

Thus by combining Equation iv & Equation vi we get;
Mm - Nm = E....... Equation iv
Nn + Mn = 0 ........ Equation vi

Now, let us solve simultaneous equation to obtain the values of M & N where m and n are both constants.

In Equation vi : Nn = -Mn

Thus N = -M, Since n and n are constants. And any valuable multiplied by constant, the answer will be constant.

So let constant M(n) be k .

Therefore the value of N will be - k.

Substitute the value of N = -k into Equation iv
Mm - Nm = E;

We get Mm + km = E
Then Mm = E - km
Then M = (E - km)/m

So let constant (E - km)/m be k .

Therefore the value of M will be k.

Thus the value of y = k(sinwt) - k(coswt)

Since the value of y = y1 + y2. The overall value of y will be given as

y = D/C + k2(sinwt) - k1(coswt) where C, D, k1 and k2 are both cosntants

duh

 

[Proved]

Dah!... Umenikumbusha kitambo Sana, Advanced mathematics. Tamu Sana hizo differential equations ukizielewa.

Hesabu tamu sana kusema kweli.

 

[Charles Mandela]

Muda wa kusolve hii mambo nishaingiza sh ngapi dukan kwa mama godiii

kilicho akilini kitumie

The clock is not ticking backward!

 

[Teknocrat]

Darius na simplemind ishu iko hivi:

Ay" + By' + Cy = D + E(sinwt).

This is a form of equation where A, B, C, D, E and w are constants and t is a parameter.

Now separate the equation:
Ay" + By' + Cy = D........Equation i

Ay" + By' + Cy = E(sinwt).........Equation ii

Then;

For Equation i
Ay" + By' + Cy = D is in form of f(X) = M ,where M is a constant and f(X) = y. Thus y = M

So if y = M; Then y' = 0 and y" = 0. Since we differentiating constant number, all derivatives become zero ( 0 ).

Then substitute the values above in Equation i

From: Ay'' + By' + Cy = D where y = M; y' = 0 and y" = 0

Then: A(0) + B(0) + C(M) = D

Then: C(M) = D , thus the value of M = D/C but M = f(x) = y

Thus the value of y = D/C


For Equation ii


Ay'' + By' + Cy = E(sinwt) is in form of f(X) = M(sinwt) + N(coswt) and f(X) = y.

Thus y = M(sinwt) + N(coswt)

So if, y = M(sinwt) + N(coswt)

Then y' =Mw(coswt) - Nw(sinwt)

Then y'' = -Mw(sinwt) - Nw(coswt)

Substitute the values above into Equation ii

A(-Mw(sinwt) - Nw(coswt)) + B(Mw(coswt) - Nw(sinwt)) + C(M(sinwt) + N(coswt)) = E(sinwt)
-AMw(sinwt) - ANw(coswt) + BMw(coswt) - BNw(sinwt) + CM(sinwt) + CN(coswt) = E(sinwt) + 0(coswt)

By comparing the values of Sine and Cosine:

For Sine

-AMw(sinwt) - BNw(sinwt) + CM(sinwt) = E(sinwt)...... Equation iii

We flash out (sinwt) and the equation will be: -AMw - BNw + CM = E

Then: M(C - Aw) - NBw = E ;

But (C - Aw) and Bw are both constants.

Then let (C - Aw) be m and Bw be m. Since all values are both constant.

Thus Mm - Nm = E....... Equation iv


For Cosine

- ANw(coswt) + BMw(coswt) + CN(coswt) = 0(coswt)...... Equation v

We flash out (coswt) and the equation will be:

-ANw + BMw + CN = 0

N(C - Aw) + BMw = 0

But (C - Aw) & Bw are both constants.

Then Let (C - Aw) be n and Bw be n.

Thus: Nn + Mn = 0 ..... Equation vi

Thus by combining Equation iv & Equation vi we get;
Mm - Nm = E....... Equation iv
Nn + Mn = 0 ........ Equation vi

Now, let us solve simultaneous equation to obtain the values of M & N where m and n are both constants.

In Equation vi : Nn = -Mn

Thus N = -M, Since n and n are constants. And any valuable multiplied by constant, the answer will be constant.

So let constant M(n) be k .

Therefore the value of N will be - k.

Substitute the value of N = -k into Equation iv
Mm - Nm = E;

We get Mm + km = E
Then Mm = E - km
Then M = (E - km)/m

So let constant (E - km)/m be k .

Therefore the value of M will be k.

Thus the value of y = k(sinwt) - k(coswt)

Since the value of y = y1 + y2. The overall value of y will be given as

y = D/C + k2(sinwt) - k1(coswt) where C, D, k1 and k2 are both cosntants

Duuhh Mkuu nimekubali

Katika ya kuafutilia mtiririko huu nimepiga miayo kadhaa
Wewe sio mchezo

 

[mathsjery]

Pure mathematics by c.j tranter

 

[Auz]

Darius na simplemind ishu iko hivi:

Ay" + By' + Cy = D + E(sinwt).

This is a form of equation where A, B, C, D, E and w are constants and t is a parameter.

Now separate the equation:
Ay" + By' + Cy = D........Equation i

Ay" + By' + Cy = E(sinwt).........Equation ii

Then;

For Equation i
Ay" + By' + Cy = D is in form of f(X) = M ,where M is a constant and f(X) = y. Thus y = M

So if y = M; Then y' = 0 and y" = 0. Since we differentiating constant number, all derivatives become zero ( 0 ).

Then substitute the values above in Equation i

From: Ay'' + By' + Cy = D where y = M; y' = 0 and y" = 0

Then: A(0) + B(0) + C(M) = D

Then: C(M) = D , thus the value of M = D/C but M = f(x) = y

Thus the value of y = D/C


For Equation ii


Ay'' + By' + Cy = E(sinwt) is in form of f(X) = M(sinwt) + N(coswt) and f(X) = y.

Thus y = M(sinwt) + N(coswt)

So if, y = M(sinwt) + N(coswt)

Then y' =Mw(coswt) - Nw(sinwt)

Then y'' = -Mw(sinwt) - Nw(coswt)

Substitute the values above into Equation ii

A(-Mw(sinwt) - Nw(coswt)) + B(Mw(coswt) - Nw(sinwt)) + C(M(sinwt) + N(coswt)) = E(sinwt)
-AMw(sinwt) - ANw(coswt) + BMw(coswt) - BNw(sinwt) + CM(sinwt) + CN(coswt) = E(sinwt) + 0(coswt)

By comparing the values of Sine and Cosine:

For Sine

-AMw(sinwt) - BNw(sinwt) + CM(sinwt) = E(sinwt)...... Equation iii

We flash out (sinwt) and the equation will be: -AMw - BNw + CM = E

Then: M(C - Aw) - NBw = E ;

But (C - Aw) and Bw are both constants.

Then let (C - Aw) be m and Bw be m. Since all values are both constant.

Thus Mm - Nm = E....... Equation iv


For Cosine

- ANw(coswt) + BMw(coswt) + CN(coswt) = 0(coswt)...... Equation v

We flash out (coswt) and the equation will be:

-ANw + BMw + CN = 0

N(C - Aw) + BMw = 0

But (C - Aw) & Bw are both constants.

Then Let (C - Aw) be n and Bw be n.

Thus: Nn + Mn = 0 ..... Equation vi

Thus by combining Equation iv & Equation vi we get;
Mm - Nm = E....... Equation iv
Nn + Mn = 0 ........ Equation vi

Now, let us solve simultaneous equation to obtain the values of M & N where m and n are both constants.

In Equation vi : Nn = -Mn

Thus N = -M, Since n and n are constants. And any valuable multiplied by constant, the answer will be constant.

So let constant M(n) be k .

Therefore the value of N will be - k.

Substitute the value of N = -k into Equation iv
Mm - Nm = E;

We get Mm + km = E
Then Mm = E - km
Then M = (E - km)/m

So let constant (E - km)/m be k .

Therefore the value of M will be k.

Thus the value of y = k(sinwt) - k(coswt)

Since the value of y = y1 + y2. The overall value of y will be given as

y = D/C + k2(sinwt) - k1(coswt) where C, D, k1 and k2 are both cosntants

Out of curiosity, What are the application of these things in real life?

 

[Charles Mandela]

Out of curiosity, What are the application of these things in real life?

Derivation of formulas

 

[Auz]

Derivation of formulas

Formulas for?

 

[share]

Darius na simplemind ishu iko hivi:

Ay" + By' + Cy = D + E(sinwt).

This is a form of equation where A, B, C, D, E and w are constants and t is a parameter.

Now separate the equation:
Ay" + By' + Cy = D........Equation i

Ay" + By' + Cy = E(sinwt).........Equation ii

Then;

For Equation i
Ay" + By' + Cy = D is in form of f(X) = M ,where M is a constant and f(X) = y. Thus y = M

So if y = M; Then y' = 0 and y" = 0. Since we differentiating constant number, all derivatives become zero ( 0 ).

Then substitute the values above in Equation i

From: Ay'' + By' + Cy = D where y = M; y' = 0 and y" = 0

Then: A(0) + B(0) + C(M) = D

Then: C(M) = D , thus the value of M = D/C but M = f(x) = y

Thus the value of y = D/C


For Equation ii


Ay'' + By' + Cy = E(sinwt) is in form of f(X) = M(sinwt) + N(coswt) and f(X) = y.

Thus y = M(sinwt) + N(coswt)

So if, y = M(sinwt) + N(coswt)

Then y' =Mw(coswt) - Nw(sinwt)

Then y'' = -Mw(sinwt) - Nw(coswt)

Substitute the values above into Equation ii

A(-Mw(sinwt) - Nw(coswt)) + B(Mw(coswt) - Nw(sinwt)) + C(M(sinwt) + N(coswt)) = E(sinwt)
-AMw(sinwt) - ANw(coswt) + BMw(coswt) - BNw(sinwt) + CM(sinwt) + CN(coswt) = E(sinwt) + 0(coswt)

By comparing the values of Sine and Cosine:

For Sine

-AMw(sinwt) - BNw(sinwt) + CM(sinwt) = E(sinwt)...... Equation iii

We flash out (sinwt) and the equation will be: -AMw - BNw + CM = E

Then: M(C - Aw) - NBw = E ;

But (C - Aw) and Bw are both constants.

Then let (C - Aw) be m and Bw be m. Since all values are both constant.

Thus Mm - Nm = E....... Equation iv


For Cosine

- ANw(coswt) + BMw(coswt) + CN(coswt) = 0(coswt)...... Equation v

We flash out (coswt) and the equation will be:

-ANw + BMw + CN = 0

N(C - Aw) + BMw = 0

But (C - Aw) & Bw are both constants.

Then Let (C - Aw) be n and Bw be n.

Thus: Nn + Mn = 0 ..... Equation vi

Thus by combining Equation iv & Equation vi we get;
Mm - Nm = E....... Equation iv
Nn + Mn = 0 ........ Equation vi

Now, let us solve simultaneous equation to obtain the values of M & N where m and n are both constants.

In Equation vi : Nn = -Mn

Thus N = -M, Since n and n are constants. And any valuable multiplied by constant, the answer will be constant.

So let constant M(n) be k .

Therefore the value of N will be - k.

Substitute the value of N = -k into Equation iv
Mm - Nm = E;

We get Mm + km = E
Then Mm = E - km
Then M = (E - km)/m

So let constant (E - km)/m be k .

Therefore the value of M will be k.

Thus the value of y = k(sinwt) - k(coswt)

Since the value of y = y1 + y2. The overall value of y will be given as

y = D/C + k2(sinwt) - k1(coswt) where C, D, k1 and k2 are both cosntants

Kwahiyo, yote hayo jibu lake ni YEYE LISSU 2020. Nimekukubali.

 

[Teknocrat]

Derivation of formulas

Formulas for?

How do populations grow? How do viruses spread?

Waache watu na fani zao
Sema tu bongo, kama ingekuwa kwa Weupe, kila kitu formula tu kwa kwenda mbele

how mathematical modeling can solve real-world problems.
👉Mathematical model - TU Delft OCW

 

[Auz]

How do populations grow? How do viruses spread?

Waache watu na fani zao
Sema tu bongo, kama ingekuwa kwa Weupe, kila kitu formula tu kwa kwenda mbele

how mathematical modeling can solve real-world problems.
Mathematical model - TU Delft OCW

Shukran kwa kunifungua macho!

 

[Teknocrat]

cc Charles Mandela, Makanyaga

wakali hii hebu saidia napelekeshwa na mpwa wangu

 

[Charles Mandela]

cc Charles Mandela, Makanyaga

wakali hii hebu saidia napelekeshwa na mpwa wangu


View attachment 1609038



View attachment 1609056

Haya ndio maswali sasa! Ngoja tuone wengine wanasemaje.

 

[Teknocrat]

Duuh kimya wakali wa maths
ok nitarudi na mrejesho nimejaribu mwenyewe baada ya kurudi university of YouTube