Question

1. $t^{2}\frac{dy}{dt}+y^{2}=ty$

Explanation:

Step1: Rewrite the differential equation

First, rewrite the given equation $t^{2}\frac{dy}{dt}+y^{2}=ty$ as $\frac{dy}{dt}=\frac{ty - y^{2}}{t^{2}}$. This is a homogeneous - type differential equation. Let $y = vt$, then $\frac{dy}{dt}=v + t\frac{dv}{dt}$.

Step2: Substitute $y = vt$ into the equation

Substitute $y = vt$ into $\frac{dy}{dt}=\frac{ty - y^{2}}{t^{2}}$. We get $v + t\frac{dv}{dt}=\frac{t\cdot vt-(vt)^{2}}{t^{2}}=\frac{v t^{2}-v^{2}t^{2}}{t^{2}}=v - v^{2}$.

Step3: Separate the variables

Rearrange the equation $v + t\frac{dv}{dt}=v - v^{2}$ to get $t\frac{dv}{dt}=-v^{2}$. Then separate the variables: $\frac{dv}{v^{2}}=-\frac{dt}{t}$.

Step4: Integrate both sides

Integrate $\int\frac{dv}{v^{2}}=\int-\frac{dt}{t}$. We know that $\int v^{-2}dv=-\frac{1}{v}+C_1$ and $\int-\frac{dt}{t}=-\ln|t| + C_2$. So, $-\frac{1}{v}=-\ln|t|+C$.

Step5: Substitute back $v=\frac{y}{t}$

Substitute $v = \frac{y}{t}$ into $-\frac{1}{v}=-\ln|t|+C$. We get $-\frac{t}{y}=-\ln|t|+C$, or $\frac{t}{y}=\ln|t| - C$. Then $y=\frac{t}{\ln|t|+C}$ (where $C$ is an arbitrary constant).

Answer:

$y = \frac{t}{\ln|t|+C}$