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^{2}}\right)…

Explanation:

Step1: Check exactness

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 $\frac{\partial M}{\partial y}=-\frac{(t^{2}+y^{2}) - y(2y)}{(t^{2}+y^{2})^{2}}=-\frac{t^{2}-y^{2}}{(t^{2}+y^{2})^{2}}$ and $\frac{\partial N}{\partial t}=\frac{(t^{2}+y^{2})-t(2t)}{(t^{2}+y^{2})^{2}}=-\frac{t^{2}-y^{2}}{(t^{2}+y^{2})^{2}}$. Since $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial t}$, the differential equation is exact.

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

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

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 $N = ye^{y}+\frac{t}{t^{2}+y^{2}}$, then $-\frac{t}{t^{2}+y^{2}}+h^{\prime}(y)=ye^{y}+\frac{t}{t^{2}+y^{2}}$. So $h^{\prime}(y)=ye^{y}$.

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

Using integration - by - parts with $u = y$, $dv=e^{y}dy$, $du = dy$, $v = e^{y}$, we have $\int ye^{y}dy=ye^{y}-\int e^{y}dy=ye^{y}-e^{y}+C$.

Step5: Write the general solution

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

Answer:

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