Question

verify that y=(4e^3x) -2 is an explicit solution of differential equation y'-3y=6

Answer 1

use the variable separation to solve this D E

Answer 2

can you try to solve it and explain more about it?

Answer 3

i think that's the right solution

Answer 4

\[\frac{ dy }{ dx }=6+3y\]
\[\frac{ dy }{ 6+3y }=dx\]
\[\int\limits_{}^{}\frac{ dy }{ 6+3y }=\int\limits_{}^{}dx\]
\[\frac{ 1 }{ 3 }\ln(6+3y)+C _{1}=x+C _{2}\]

Answer 5

\[\ln(6+3y)=3x+C\]
\[e ^{\ln(6+3y)}=e ^{3x+C}\]
\[6+3y=e ^{3x}e ^{C}\]
\[3y=Ce ^{3x}-6\]
\[y=Ce ^{3x}-2\]

Answer 6

I think you need to know an intial value y_0 to find C = 4.

Answer 7

This question does not ask you to solve the DE,
it asks to verify that a given solution solves the DE.

Answer 8

The function \[y(x)=4e^{3x} -2\]
its derivative\[y'(x) = 12e^{3x}\]
The differential equation
\[\qquad y'-3y\qquad \quad =6\]
Plugging the function and its derivative into the DE\[\begin{align}
[12e^{3x}]-3[4e^{3x} -2]&=6\\
12e^{3x}-12e^{3x} +6&=6\\
6&=6\\
0&=0
\end{align}\](a true statement)

Hence \(y=4e^{3x} -2\) is an explicit solution to the differential equation.