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- Quiz
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- MTH401 Differential Equations
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- PDF 41 KB
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- 23 Oct 2026
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Text from pages 2, 3, 4 of 7. Download the file for all 7 pages.
Question:
If y=2+xy=2+xy = 2 + x , then which of the following is true for it?
Copied: 11/25/2025, 10:40:34 AM
Question:
If y=c1ex+c2e4xy=c1ex+c2e4xy = {c_1}{e^x} + {c_2}{e^{4x}} is the complementary solution of �y 2 2"�5�y 2�+�4�y�=�8�e�x�y 3"�5�y 2�+�4�y�=�8�e�x�y�'�'� �-� �5�y�'� �+� �4�y� �=� �8�{�e�^�x�}� �,� �t�h�e�n� �t�h�e� �g�e�n�e�r�a�l� �f�o�r�m� �o ypyp{y_p}=_________.
A. None of these B. AxexAxexAx{e^x} C. AexAexA{e^x} D. (Ax+B)e4x(Ax+B)e4x(Ax + B){e^{4x}}
Copied: 11/25/2025, 10:40:48 AM
Question:
g(x)=5x+7,thentheassumedparticularsolutionypcouldbe_________.Considerthenon - homogeneouslineardifferential �e�q�u�a�t�i�o�n�,�a�n�y�(�n�)�+�a�n"�1�y�(�n"�1�)�+�.�.�.�+�a�1�y 2�+�a�0�y�=�g�(�x�)�.� �I�f�t�h�e�i�n�p�u�t�f�u�n�c�t�i�o�n�,�g�(�x�)�=�5�x�+�7�,�t�h�e�n�t�h�e�a�s�s�u�m�e�d� ouldbe_________.\begin{gathered}
{\text{Consider}}\,\,{\text{the}}\,\,{\text{non -
homogeneous}}\,\,{\text{linear}}\,\,{\text{differential}}\,\,{\text{equation,}}\,\,{a_n}{y^{(n)}} + {a_{n - 1}}{y^{(n - 1)}} + ... + {a_1}y' + {a_0}y = g(x).\,\, \hfill \\\ If\,\,{\text{the}}\,\,{\text{input}}\,\,{\text{function,}}\,\,g(x) = 5x + 7,\,\,{\text{then}}\,\, {\text{the}}\,\,{\text{assumed}}\,\,{\text{particular}}\,\,{\text{solution}}\,\,{y_p}\,\,{\text{could}}\,\,{\text{be}}\,\,{\text{____
_____}}{\text{.}}\, \hfill \\\\
\end{gathered}
A. AAA B. Ax2+Bx+CAx2+Bx+CA{x^2} + Bx + C C. Ax3+Bx2+Cx+DAx3+Bx2+Cx+DA{x^3} + B{x^2} + Cx + D D. Ax+BAx+BAx + B
Copied: 11/25/2025, 10:41:01 AM
Question:
�g�(�x�)�=�3�x�2"�2�,�t�h�e�n�t�h�e�a�s�s�u�m�e�d�p�a�r�t�i�c�u�l�a�r�s�o�l�u�t�i�o�n�y�p�c�o�u�l�d�b�e�_�_�_�_�_�_�_�_�_�.�C�o�n�s�i�d�e�r�t�h�e �e�q�u�a�t�i�o�n�,�a�n�y�(�n�)�+�a�n"�1�y�(�n"�1�)�+�.�.�.�+�a�1�y 2�+�a�0�y�=�g�(�x�)�.� �I�f�t�h�e�i�n�p�u�t�f�u�n�c�t�i�o�n�,�g�(�x�)�=�3�x�2"�2� couldbe_________.\begin{gathered} {\text{Consider}}\,\,{\text{the}}\,\,{\text{non - homogeneous}}\,\,{\text{linear}}\,\,{\text{differential}}\,\,{\text{equation,}}\,\,{a_n}{y^{(n)}} + {a_{n - 1}}{y^{(n - 1)}} + ... + {a_1}y' + {a_0}y = g(x).\,\, \hfill \\\ If\,\,{\text{the}}\,\,{\text{input}}\,\,{\text{function,}}\,\,g(x) = 3{x^2} - 2,\,\,{\text{then} }\,\,{\text{the}}\,\,{\text{assumed}}\,\,{\text{particular}}\,\,{\text{solution}}\,\,{y_p}\,\,{\text{could}}\,\,{\text{be}}\,\,{\text{__ _______}}{\text{.}}\, \hfill \\\\ \end{gathered}
A. AAA B. Ax2+Bx+CAx2+Bx+CA{x^2} + Bx + C C. Ax+BAx+BAx + B D. Ax3+Bx2+Cx+DAx3+Bx2+Cx+DA{x^3} + B{x^2} + Cx + D
Copied: 11/25/2025, 10:41:32 AM
Question:
In the population growth model, the solution of the differential equation is given by ______.
A. A(t)=A0/2A(t)=A0/2A(t) = {A_0}/2 B. A(t)=A0ektA(t)=A0ektA(t) = {A_0}{e^{kt}} C. A(t)=kektA(t)=kektA(t) = k{e^{kt}} D. A(t)=A0/2A(t)=A0/2A(t) = {A_0}/2
Copied: 11/25/2025, 10:41:49 AM
Question:
�A�=#¡#£#¢�1�1�/�4�1�/�2�1�/�4�1�/�3�1�/�4�1�/�2�1�/�4�1�/�5#¤#¦#¥�A�=�[�1�1�/�4�1�/�2�1�/�4�1�/�3�1�/�4�1�/�2�1�/�4�1�/�5�]�A� �=� �\�l�e�f�t�[� � 1&{1/4}&{1/2}\\{1/4}&{1/3}&{1/4}\\{1/2}&{1/4}&{1/5}
�a�n�d� �‘�t�h�e�t�a� ¸� �=�0�.�4�4�8�0�,� �t�h�e� �o�r�t�h�o�g�o�n�a�l� �m�a�t�r�i�x� �S�1� �w�i�l�l� �b�e� �g�i�v�e�n� �b�y
�A�.� �S�1�=#¡#£#¢�c�o�s�0�.�4�4�8�0�0�s�i�n�0�.�4�4�8�0�0�1�0"�s�i�n�0�.�4�4�8�0�0�c�o�s�0�.�4�4�8�0#¤#¦#¥�S�1�=�[�c�o�s a�0�.�4�4�8 �o�s a�0�.�4�4�8�0�]�{�S�_�1�}� �=� �\�l�e�f�t�[� �{�\�b�e�g�i�n�{�a�r�r�a�y�}�{�*�{�2�0�}�{�c�}�} {\cos 0.4480}&0&{ - \sin 0.4480}\\0&1&0\\{\sin 0.4480}&0&{\cos 0.4480}
�B�.� �S�1�=#¡#£#¢�c�o�s�0�.�4�4�8�0�s�i�n�0�.�4�4�8�0�0"�s�i�n�0�.�4�4�8�0�c�o�s�0�.�4�4�8�0�0�0�0�1#¤#¦#¥�S�1�=�[�c�o�s a�0�.�4�4�8 44800001]{S_1} = \left[ {\begin{array}{*{20}{c}} {\cos 0.4480}&{ - \sin 0.4480}&0\\{\sin 0.4480}&{\cos 0.4480}&0\\0&0&1
�C�.� �S�1�=#¡#£#¢�1�0�0�0�c�o�s�0�.�4�4�8�0�s�i�n�0�.�4�4�8�0�0"�s�i�n�0�.�4�4�8�0�c�o�s�0�.�4�4�8�0#¤#¦#¥�S�1�=�[�1�0�0�0�c�o�s a�0 �o�s a�0�.�4�4�8�0�]�{�S�_�1�}� �=� �\�l�e�f�t�[� �{�\�b�e�g�i�n�{�a�r�r�a�y�}�{�*�{�2�0�}�{�c�}�} 1&0&0\\0&{\cos 0.4480}&{ - \sin 0.4480}\\0&{\sin 0.4480}&{\cos 0.4480}
�D�.� �S�1�=#¡#£#¢�c�o�s�0�.�4�4�8�0�0�0�0"�s�i�n�0�.�4�4�8�0�0�0�0�1#¤#¦#¥�S�1�=�[�c�o�s a�0�.�4�4�8�0�0�0�0"�s�i�n a�0�.�4�4�8�0�0 {\begin{array}{*{20}{c}}
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