Question

solve the given differential equation by using an appropriate substitution. the de is a bernoulli equation.
\\(t^{2}\frac{dy}{dt}+y^{2}=ty\\)

Explanation:

Step1: Rewrite the Bernoulli equation

The given differential equation is $\frac{dy}{dt}+y^{2}=ty$. Divide by $y^{2}$ (assuming $y
eq0$) to get $y^{-2}\frac{dy}{dt}+ 1 = ty^{-1}$. Let $z = y^{-1}$, then $\frac{dz}{dt}=-y^{-2}\frac{dy}{dt}$. The equation becomes $-\frac{dz}{dt}+1 = tz$, or $\frac{dz}{dt}+tz = 1$.

Step2: Find the integrating - factor

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

Step3: Multiply the equation by the integrating factor

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

Step4: Integrate both sides

Integrate $\frac{d}{dt}(e^{\frac{t^{2}}{2}}z)=e^{\frac{t^{2}}{2}}$ with respect to $t$. $\int\frac{d}{dt}(e^{\frac{t^{2}}{2}}z)dt=\int e^{\frac{t^{2}}{2}}dt$. The left - hand side is $e^{\frac{t^{2}}{2}}z$. The integral $\int e^{\frac{t^{2}}{2}}dt$ cannot be expressed in terms of elementary functions. Let $\int e^{\frac{t^{2}}{2}}dt = \sqrt{\pi}\text{erfi}(\frac{t}{\sqrt{2}})+C$ (where $\text{erfi}(x)$ is the imaginary error function). So $e^{\frac{t^{2}}{2}}z=\sqrt{\pi}\text{erfi}(\frac{t}{\sqrt{2}})+C$.

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

Since $z = y^{-1}$, we have $y=\frac{1}{z}$. So $y=\frac{1}{e^{-\frac{t^{2}}{2}}(\sqrt{\pi}\text{erfi}(\frac{t}{\sqrt{2}})+C)}$. Also, $y = 0$ is a singular solution of the original Bernoulli equation.

Answer:

$y = 0$ and $y=\frac{1}{e^{-\frac{t^{2}}{2}}(\sqrt{\pi}\text{erfi}(\frac{t}{\sqrt{2}})+C)}$