20. $\\left(\\frac{1}{t}+\\frac{1}{t^{2}}-\\frac{y}{t^{2}+y^{2}}\\right)dt + \\left(ye^{y}+\\frac{t}{t^{2}+y^…

Explanation:

Step1: Check if it's an exact differential equation

Let $M(t,y)=\frac{1}{t}+\frac{1}{t^{2}}-\frac{y}{t^{2}+y^{2}}$ and $N(t,y)=ye^{y}+\frac{t}{t^{2}+y^{2}}$. Calculate $\frac{\partial M}{\partial y}$ and $\frac{\partial N}{\partial t}$. $\frac{\partial M}{\partial y}=-\frac{1}{t^{2}+y^{2}}+\frac{2y^{2}}{(t^{2}+y^{2})^{2}}$ and $\frac{\partial N}{\partial t}=\frac{1}{t^{2}+y^{2}}-\frac{2t^{2}}{(t^{2}+y^{2})^{2}}$. Since $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial t}$, it is an exact differential equation.

Step2: Integrate $M$ with respect to $t$

$\int M(t,y)dt=\int\left(\frac{1}{t}+\frac{1}{t^{2}}-\frac{y}{t^{2}+y^{2}}\right)dt=\ln|t|-\frac{1}{t}- \arctan\left(\frac{t}{y}\right)+h(y)$

Step3: Differentiate the result with respect to $y$ and equate to $N$

$\frac{\partial}{\partial y}\left(\ln|t|-\frac{1}{t}- \arctan\left(\frac{t}{y}\right)+h(y)\right)=\frac{t}{t^{2}+y^{2}}+h^{\prime}(y)$. Since $N(t,y)=ye^{y}+\frac{t}{t^{2}+y^{2}}$, then $h^{\prime}(y)=ye^{y}$.

Step4: Integrate $h^{\prime}(y)$ to find $h(y)$

Using integration - by - parts $\int ye^{y}dy=(y - 1)e^{y}+C$. So $h(y)=(y - 1)e^{y}+C$.

Step5: Write the general solution

The general solution of the exact differential equation is $\ln|t|-\frac{1}{t}-\arctan\left(\frac{t}{y}\right)+(y - 1)e^{y}=C$.

Answer:

$\ln|t|-\frac{1}{t}-\arctan\left(\frac{t}{y}\right)+(y - 1)e^{y}=C$