Hi, I'm having trouble finding the general solution to the following linear ODE: (dy/dt) + y = (t^3) + sin(3t). I know that the general solution to the homogeneous equation is ke^(-t), but I'm not sure where to begin to find a particular solution to the nonhomogeneous equation. Thank you all in advance.
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General solution to a linear ODE
- Thread starter Gladius
- Start date
Hello, Gladius!
Have you tried Undetermined Coefficients?
\(\displaystyle \text{We conjecture that the solution is of the form: }\:y \;=\;At^3 + Bt^2 + Ct + D + E\sin(3t) + F\cos(3t)\;\;[1]\)
\(\displaystyle \text{Then: }\:\frac{dy}{dt} \;=\;2At^2 + 2Bt + C + 3E\cos(3t) - 3F\sin(3t)\;\;[2]\)
\(\displaystyle \text{Add [1] and [2]:}\)
. . \(\displaystyle \frac{dy}{dt} \;+\; y \;=\;At^3 \;+\; (3A+B)t^2 \;+\; (2B+C)t \;+\; (C+D) \;+\; (E-3F)\sin(3t) \;+\; (3E+F)\cos(3t)\)
\(\displaystyle \text{Since this equals: }\:1\!\cdot\!t^3 + 0\!\cdot\!t^2 + 0\!\cdot\!t + 0 + 1\!\cdot\!\sin(3t) + 0\!\cdot\!\cos(3t)\)
. . \(\displaystyle \text{we equate the coefficients: }\;\begin{bmatrix}A &=& 1 \\ 3A + B &=&0 \\ 2B+C &=&0 \\ C+D &=&0 \\ E-3F &=&1 \\ 3E + F &=&0 \end{bmatrix}\)
. . \(\displaystyle \text{and solve the system: }\;\begin{Bmatrix}A &=&1 \\ B &=& \text{-}3 \\ C&=&6 \\ D &=&\text{-}6 \\ E &=&\frac{1}{10} \\ \\[-3mm]F &=&\text{-}\frac{3}{10} \end{Bmatrix}\)
\(\displaystyle \text{Therefore, the particular solution is: }\;y \;=\;t^3 \;-\; 3t^2 \;+\; 6t \;-\; 6 \;+\; \tfrac{1}{10}\sin(3t) \;-\; \tfrac{3}{10}\cos(3t)\)
\(\displaystyle \text{I'm having trouble finding the general solution to this linear ODE: }\:\frac{dy}{dt} + y \;=\; t^3 + \sin(3t)\)
\(\displaystyle \text{I know that the general solution to the homogeneous equation is: }\:ke^{-t}\)
\(\displaystyle \text{but I'm not sure where to begin to find a particular solution to the nonhomogeneous equation.}\)
Have you tried Undetermined Coefficients?
\(\displaystyle \text{We conjecture that the solution is of the form: }\:y \;=\;At^3 + Bt^2 + Ct + D + E\sin(3t) + F\cos(3t)\;\;[1]\)
\(\displaystyle \text{Then: }\:\frac{dy}{dt} \;=\;2At^2 + 2Bt + C + 3E\cos(3t) - 3F\sin(3t)\;\;[2]\)
\(\displaystyle \text{Add [1] and [2]:}\)
. . \(\displaystyle \frac{dy}{dt} \;+\; y \;=\;At^3 \;+\; (3A+B)t^2 \;+\; (2B+C)t \;+\; (C+D) \;+\; (E-3F)\sin(3t) \;+\; (3E+F)\cos(3t)\)
\(\displaystyle \text{Since this equals: }\:1\!\cdot\!t^3 + 0\!\cdot\!t^2 + 0\!\cdot\!t + 0 + 1\!\cdot\!\sin(3t) + 0\!\cdot\!\cos(3t)\)
. . \(\displaystyle \text{we equate the coefficients: }\;\begin{bmatrix}A &=& 1 \\ 3A + B &=&0 \\ 2B+C &=&0 \\ C+D &=&0 \\ E-3F &=&1 \\ 3E + F &=&0 \end{bmatrix}\)
. . \(\displaystyle \text{and solve the system: }\;\begin{Bmatrix}A &=&1 \\ B &=& \text{-}3 \\ C&=&6 \\ D &=&\text{-}6 \\ E &=&\frac{1}{10} \\ \\[-3mm]F &=&\text{-}\frac{3}{10} \end{Bmatrix}\)
\(\displaystyle \text{Therefore, the particular solution is: }\;y \;=\;t^3 \;-\; 3t^2 \;+\; 6t \;-\; 6 \;+\; \tfrac{1}{10}\sin(3t) \;-\; \tfrac{3}{10}\cos(3t)\)