## Related questions with answers

Consider the differential equation $d y / d t=t^2 y+1+y+t^2$. (a) Find its general solution by separating variables. (b) Note that this equation is also a nonhomogeneous linear equation. Find the general solution of its associated homogeneous equation. (c) Calculate the equilibrium solutions of the nonhomogeneous equation. (d) Using the Extended Linearity Principle, find the general solution of the nonhomogeneous equation. Compare your result to the one you obtained in part (a).

Solution

VerifiedSeparating the variables and then integrating both sides:

$\begin{align*} \dfrac{dy}{dt}=t^{2}y+1+y+t^{2}&=\left(1+t^{2}\right)\left(y+1\right)\\ \quad \longrightarrow \quad \dfrac{1}{y+1} \; dy &= \left(1+t^{2} \right) \; dt\\ \int \dfrac{1}{y+1} \; dy &= \int \left(1+t^{2} \right) \;dt\\ \ln\left| y+1 \right| &=t+\dfrac{t^{3}}{3} +k_{1}\\ \text{where $k_{1}$ is an arbitrary constant.}\\ \left|y+1\right| &= e^{t+\frac{t^{3}}{3}+k_{1}}\\ \left|y+1\right| &= e^{k_{1}}e^{t+\frac{t^{3}}{3}}\\ \left|y+1\right| &=k_{2}e^{t+\frac{t^{3}}{3}}\\ \end{align*}$

here $e^{k_{1}}=k_{2}$ where $k_{2}$ is an arbitrary positive constant.

To get rid of the absolute sign, we use the properties of the absolute value:

$\begin{align*}y+1 &= \pm k_{2} e^{t+\frac{t^{3}}{3}}\\ y(t) &= ke^{t+\frac{t^{3}}{3}}-1\\ \text{here $\pm k_{2}=k$ where $k$ is an arbitrary constant.} \end{align*}$

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