solve the given differential equation by using an appropriate substitution. the de is a bernoulli…

Explanation:

Step1: Rewrite the Bernoulli equation

The given equation is $\frac{dy}{dt}+y^{2}=ty$. Divide by $y^{2}$ (assuming $y\neq0$) to get $y^{- 2}\frac{dy}{dt}+1 = ty^{-1}$. Let $u = y^{-1}$, then $\frac{du}{dt}=-y^{-2}\frac{dy}{dt}$. The equation becomes $-\frac{du}{dt}+1 = tu$, or $\frac{du}{dt}+tu = 1$.

Step2: Find the integrating factor

The integrating factor for the first - order linear differential equation $\frac{du}{dt}+P(t)u = Q(t)$ (here $P(t)=t$ and $Q(t) = 1$) is $\mu(t)=e^{\int P(t)dt}=e^{\int tdt}=e^{\frac{t^{2}}{2}}$.

Step3: Multiply the equation by the integrating factor

Multiply $\frac{du}{dt}+tu = 1$ by $e^{\frac{t^{2}}{2}}$, we get $e^{\frac{t^{2}}{2}}\frac{du}{dt}+te^{\frac{t^{2}}{2}}u=e^{\frac{t^{2}}{2}}$. The left - hand side is the derivative of the product $e^{\frac{t^{2}}{2}}u$ with respect to $t$, i.e., $\frac{d}{dt}(e^{\frac{t^{2}}{2}}u)=e^{\frac{t^{2}}{2}}$.

Step4: Integrate both sides

Integrate $\frac{d}{dt}(e^{\frac{t^{2}}{2}}u)=e^{\frac{t^{2}}{2}}$ with respect to $t$. $\int\frac{d}{dt}(e^{\frac{t^{2}}{2}}u)dt=\int e^{\frac{t^{2}}{2}}dt$. Let $I=\int e^{\frac{t^{2}}{2}}dt$. We know that $\int e^{\frac{t^{2}}{2}}dt=\sqrt{\pi}\text{erfi}(\frac{t}{\sqrt{2}})+C$ (where $\text{erfi}(x)$ is the imaginary error function). So $e^{\frac{t^{2}}{2}}u=\int e^{\frac{t^{2}}{2}}dt + C$.

Step5: Solve for $u$ and then $y$

$u = e^{-\frac{t^{2}}{2}}\left(\int e^{\frac{t^{2}}{2}}dt + C\right)$. Since $u = y^{-1}$, then $y=\frac{1}{e^{-\frac{t^{2}}{2}}\left(\int e^{\frac{t^{2}}{2}}dt + C\right)}$.

Answer:

$y=\frac{1}{e^{-\frac{t^{2}}{2}}\left(\int e^{\frac{t^{2}}{2}}dt + C\right)}$