Question 9 The general solution for the DE y + 4y=tan(2x) is given by cy=c;cos(2x) + czsin(2x) + = sin(2 x)In|sec(2x) + tan(2x)|. Oc. O e. None of these dy= c,cos(2x) + czsin(2x) – ÷ sin(2x)In|sec(2x) + tan(2 x)|. O b. y= c;cos(2x) + czsin(2x) – cos(2x)In|sec(2x) + tan(2x)|. a y= c¿cos(2x) + czsin(2x)+ ÷cos(2×)in|sec(2x) + tan(2x)|

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.2: Trigonometric Equations
Problem 75E
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Question 9
The general solution for the DE y + 4y=tan(2x) is given by
Ocy=ccos(2x)+ c,sin(2x) + =sin(2x)In|sec(2x) + tan(2x)|.
O e. None of these
dy= ccos(2x) + czsin(2x) – = sin(2x)In|sec(2x) + tan(2 x}| .
4
by= cqcos(2x) + czsin(2x) – cos(2x)In|sec(2x) + tan(2x)|.
Ma y= c,cos(2x) + czsin(2x) + ÷cos(2×)ln|sec(2x) + tan(2x)|
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Transcribed Image Text:A Moving to another question will save this response. Question 9 The general solution for the DE y + 4y=tan(2x) is given by Ocy=ccos(2x)+ c,sin(2x) + =sin(2x)In|sec(2x) + tan(2x)|. O e. None of these dy= ccos(2x) + czsin(2x) – = sin(2x)In|sec(2x) + tan(2 x}| . 4 by= cqcos(2x) + czsin(2x) – cos(2x)In|sec(2x) + tan(2x)|. Ma y= c,cos(2x) + czsin(2x) + ÷cos(2×)ln|sec(2x) + tan(2x)| Moving to another question will save this response.
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