To prove: If a 0 = 0 and a 1 ≠ 0 in a n y ( n ) ( x ) + a n − 1 y ( n − 1 ) ( x ) + ... + a 0 y ( x ) = f ( x ) and f ( x ) = b m x m + b m − 1 x m − 1 + ... + b 1 x + b 0 , then the particular solution is of the form y p ( x ) = x ( B m x m + B m − 1 x m − 1 + ... + B 1 x + B 0 ) .

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Answer 1

Question

Chapter 6.3, Problem 35E

To determine

To prove:

If a0=0 and a1≠0 in any(n)(x)+an−1y(n−1)(x)+...+a0y(x)=f(x) and f(x)=bmxm+bm−1xm−1+...+b1x+b0, then the particular solution is of the form yp(x)=x(Bmxm+Bm−1xm−1+...+B1x+B0).

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Answer 2

Question

Chapter 6.3, Problem 35E

To determine

To prove:

If a0=0 and a1≠0 in any(n)(x)+an−1y(n−1)(x)+...+a0y(x)=f(x) and f(x)=bmxm+bm−1xm−1+...+b1x+b0, then the particular solution is of the form yp(x)=x(Bmxm+Bm−1xm−1+...+B1x+B0).

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Answer 3

Question

Chapter 6.3, Problem 35E

To determine

To prove:

If a0=0 and a1≠0 in any(n)(x)+an−1y(n−1)(x)+...+a0y(x)=f(x) and f(x)=bmxm+bm−1xm−1+...+b1x+b0, then the particular solution is of the form yp(x)=x(Bmxm+Bm−1xm−1+...+B1x+B0).

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Answer 4

Question

Chapter 6.3, Problem 35E

To determine

To prove:

If a0=0 and a1≠0 in any(n)(x)+an−1y(n−1)(x)+...+a0y(x)=f(x) and f(x)=bmxm+bm−1xm−1+...+b1x+b0, then the particular solution is of the form yp(x)=x(Bmxm+Bm−1xm−1+...+B1x+B0).

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Answer 5

Chapter 6.3, Problem 35E

To determine

To prove:

If a0=0 and a1≠0 in any(n)(x)+an−1y(n−1)(x)+...+a0y(x)=f(x) and f(x)=bmxm+bm−1xm−1+...+b1x+b0, then the particular solution is of the form yp(x)=x(Bmxm+Bm−1xm−1+...+B1x+B0).

Answer 6

Chapter 6.3, Problem 35E

To determine

To prove:

If a0=0 and a1≠0 in any(n)(x)+an−1y(n−1)(x)+...+a0y(x)=f(x) and f(x)=bmxm+bm−1xm−1+...+b1x+b0, then the particular solution is of the form yp(x)=x(Bmxm+Bm−1xm−1+...+B1x+B0).

Answer 7

Chapter 6.3, Problem 35E

To determine

To prove:

If a0=0 and a1≠0 in any(n)(x)+an−1y(n−1)(x)+...+a0y(x)=f(x) and f(x)=bmxm+bm−1xm−1+...+b1x+b0, then the particular solution is of the form yp(x)=x(Bmxm+Bm−1xm−1+...+B1x+B0).

Answer 8

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Answer 9

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Answer 10

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Answer 11

Ch. 6.1 - In Problems 1-6, determine the largest interval...Ch. 6.1 - In Problems 1-6, determine the largest interval...Ch. 6.1 - In Problems 1-6, determine the largest interval...Ch. 6.1 - In Problems 1-6, determine the largest interval...Ch. 6.1 - In Problems 1-6, determine the largest interval...Ch. 6.1 - In Problems 1-6, determine the largest interval...Ch. 6.1 - Prob. 7E Ch. 6.1 - In Problems7-14, determine whether the given...Ch. 6.1 - In Problems7-14, determine whether the given...Ch. 6.1 - In Problems7-14, determine whether the given...

To prove: If a 0 = 0 and a 1 ≠ 0 in a n y ( n ) ( x ) + a n − 1 y ( n − 1 ) ( x ) + ... + a 0 y ( x ) = f ( x ) and f ( x ) = b m x m + b m − 1 x m − 1 + ... + b 1 x + b 0 , then the particular solution is of the form y p ( x ) = x ( B m x m + B m − 1 x m − 1 + ... + B 1 x + B 0 ) .

Solution Summary: The author proves that the particular solution is of the form y_p(x)=x

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To prove:

If a0=0 and a1≠0 in any(n)(x)+an−1y(n−1)(x)+...+a0y(x)=f(x) and f(x)=bmxm+bm−1xm−1+...+b1x+b0, then the particular solution is of the form yp(x)=x(Bmxm+Bm−1xm−1+...+B1x+B0).

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Chapter 6 Solutions

To prove: If a 0 = 0 and a 1 ≠ 0 in a n y ( n ) ( x ) + a n − 1 y ( n − 1 ) ( x ) + ... + a 0 y ( x ) = f ( x ) and f ( x ) = b m x m + b m − 1 x m − 1 + ... + b 1 x + b 0 , then the particular solution is of the form y p ( x ) = x ( B m x m + B m − 1 x m − 1 + ... + B 1 x + B 0 ) .

Solution Summary: The author proves that the particular solution is of the form y_p(x)=x

Concept explainers

Videos

To prove: If a0=0 and a1≠0 in any(n)(x)+an−1y(n−1)(x)+...+a0y(x)=f(x) and f(x)=bmxm+bm−1xm−1+...+b1x+b0, then the particular solution is of the form yp(x)=x(Bmxm+Bm−1xm−1+...+B1x+B0).

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Chapter 6 Solutions

To prove:

If a0=0 and a1≠0 in any(n)(x)+an−1y(n−1)(x)+...+a0y(x)=f(x) and f(x)=bmxm+bm−1xm−1+...+b1x+b0, then the particular solution is of the form yp(x)=x(Bmxm+Bm−1xm−1+...+B1x+B0).