(D³ – 4D² + 4D)y = 0 (4D² – 6D + 1)y = 0 %3D

Q: 2. (D2 - 5D + 6)y = 2 e*. 3. (D2 - 4D + 4)y = e*. |

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Q: b. (D² – 4D + 4)y = e2x

A: problem is solved using CF and PI method

Q: 4. (D² +D-1)² y =0 ,when x= y = 0, y' = y" =1, y" = -1

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Q: c2 c4-x² xe²y 2. dy dx 4-y

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Q: 1. (D²-5D+6)y = 2e

A: This is a problem of differential equation.

Q: 10. (D² – 6D + 9)y = e3x

A: Given, D2-6D+9y=e3x To solve the above differential equation.

Q: y" + 2y' – 15y = e3x %3D

A: We find particular integral by inverse operator

Q: (D² +9)y=sec3x

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Q: 1. (D² -5D³)y = 4 2. (D³ – 5D* + 7D³ +D² -8D+4)y = 0

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Q: -. (D² +4D+4)y=-x*e y =-x³e

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Q: d. ' = [ * -2 e. X' = .ジ [4 -3 え

A: This is a question of Differential Equation.

Q: D2 – 5D +6)y=3xex+2x2 , y(0)=0,y,(0)=1

A: Given differential equation is homogeneous ordinary differential equation. To solve this equation…

Q: Find EF in the trapezoid. x+7 D E 3x 8 B

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Q: d'y d³y - dx5 dx3 = 24.

A: Solution :-

Q: c. The f'(3) for y = v3x² + 9

A: c)y=3x2+9

Q: 2 d2 ; C:Z10)=3e ;osOs下

A: we will evaluate this integral by using parameterization since we have given a function in z and…

Q: (D2 – 1)y = 2e-*(1+e-2*)-2, %3D

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Q: (i) (D² – 4D+ 13) y=e2r 4x (ii) (D² + 16) y=e*

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Q: d*y + 2 dx3 d³y , d²y 8. dx4 = 0 dx2

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Q: c2 Find (4x + 3y)°dædy

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Q: (e) y(ª) – 4y "= 5x² -e2* (g) y(“) +8y '= 4

A: The given differential equation is y(4)-4y''=5x2-e2x

Q: 2 y (4- y²)³ dy

A: The objective of the question is evaluate the definite integral.

Q: (D3 + 7D2 + 19D + 13)y = 0

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Q: (D³ -3D* +4D³ -8D² +16),y = 0

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Q: 19.(D³ + 36D)²y = 0 %3D

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Q: 3. (D² — 2D + 1) у %3 е* In a

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Q: (D2 – 4D + 4)y = e3x

A: Given ODE is: (D2-4D+4)y=e3x

Q: y" – 3y + 2y = e5a

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Q: - y² dy 3y 50.

A: Simplify the expression and integrate it.

Q: 6. (D3 – D)y = -2(1 + x) + 4e* – 6e-* + 96e3x

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Q: y = 2t + t2 – 2(In t)² d.

A: y = 2t + t2 - 2(lnt)2

Q: 6. (D - 4D2 + 7D)y = 0 (D3

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Q: (D³ + 7D² + 19D + 13)y 0 when x = 0, y = 0, y' = 2, y" = -12

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Q: (D² – 6D + 9)y = 0 ; x = 0; y = 0; y' = 2 %3D %3D

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Q: (d) y" – 6y" + 11y' – 6y = e2" (5 – 4x – 3a²)

A: Given  A non homogeneous differential equation in question 2. D.

Q: メe %3D find df 4 メ=ら トメ %3D 3.

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Q: y² + 5 dy y²(y² + 25)

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Q: 3. (D5 –Dª – 2D³ +18D² –31D+15)y =0

A: Given differential equation is D5-D4-2D3+18D2-31D+15y = 0

Q: (D3−D2+4D−4)y=0 ; y(0)=2, y'(0)=7, y"(0)=-3

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Q: (D² – 5D) + 4y = 0 (D² – 9)y = 0

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Q: y = (t² – 2)². a²y 2. dx²

A: y=t2-22 and x=t2 Now, differentiate the given function with respect to t.…

Q: 6. (D² + 1)y =- 1 %3D 3

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Q: (D² – 8D + 16)y = 4e¬v2t, y(0) = -1, y'(0) = 3;

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Q: d²y ; for 9x² + y²=9 Find dx2 81x2- 2xy A y4 18x + 2y B -81

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Q: (D² + 6D + 7)(D3 – 5D² + 6D)y = %3D

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Q: 2. (D5 –D – 2D³ + 18D² –31D+15)y =0

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Q: 2. (D³ +5D² + 6D)y=24e -3x. Use two methods.

A: Here solving given DE (2) in two different method.If you want anything else kindly repost the…

Q: d*y d³y - 4· dx3 +7 d²y 14 dx4 dx2

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Q: (D4 + 6D3 + 9D2)y = 0

A: