Exact Differential Equations

Let $y=y\left(x\right)$ be the solution of the differential equation $\left(x+1\right)y^{\text{'}}-y={\mathrm{e}}^{3x}{\left(x+1\right)}^2$, with $y\left(0\right)=\frac{1}{3}$. Then, the point $x=-\frac{4}{3}$ for the curve $y=y\left(x\right)$ is

Let $y=y\left(x\right)$ be the solution of the differential equation $xdy=\left(y+x^3\cos x\right)dx$ with $y\left(\pi \right)=0,$ then $y\left(\frac{\pi }{2}\right)$ is equal to:

Let $y=y\left(x\right)$ be the solution of the differential equation $\left(\left(x+2\right)e^{\left(\frac{y+1}{x+2}\right)}+\left(y+1\right)\right)dx=\left(x+2\right)dy, y\left(1\right)=1.$ If the domain of $y=y\left(x\right)$ is an open interval $(\alpha , \beta ),$ then $|\alpha +\beta |$ is equal to ___________.

Let $y=y\left(x\right)$ be the solution of the differential equation $\frac{dy}{dx}=1+x{e}^{y-x}, -\sqrt{2}<x<\sqrt{2}, y\left(0\right)=0$, then the minimum value of $y\left(x\right), x\in \left(-\sqrt{2},\sqrt{2}\right)$ is equal to :

Let $y=y\left(x\right)$ be the solution of the differential equation $\left(x-x^3\right)dy=\left(y+y{x}^2-3x^4\right)dx,x>2$ If $y\left(3\right)=3,$ then $y\left(4\right)$ is equal to:

If the curve, $y=y\left(x\right)$ represented by the solution of the differential equation $\left(2x{y}^2-y\right)dx+x dy=0$, passes through the intersection of the lines, $2x-3y=1$ and $3x+2y=8$, then $\left|y\left(1\right)\right|$ is equal to ___ .

Let $f$ be a twice differentiable function defined on $R$ such that $f\left(0\right)=1, f^{\text{'}}\left(0\right)=2$ and $f^{\text{'}}\left(x\right)\ne 0$ for all $x\in R.$ If $\left|\begin{array}{ll}f\left(x\right)& f^{\text{'}}\left(x\right)\\ f^{\text{'}}\left(x\right)& f^{\text{'}\text{'}}\left(x\right)\end{array}\right|=0,$ for all $x\in R,$ then the value of $f\left(1\right)$ lies in the interval

The solution of the differential equation $\frac{dy}{dx}-\frac{y+3x}{{\log }_e(y+3x)}+3=0$ is
(where $C$ is a constant of integration)

A curve amongst the family of curves represented by the differential equation, $\left(x^2-y^2\right) dx+2xy dy=0$ which passes through $\left(1,1\right)$, is

If a curve $y=f\left(x\right)$ passes through the point $(1,-1)$ and satisfies the differential equation, $y \left(1+xy\right)dx=x dy,$ then $f\left(-\frac{1}{2}\right)$ is equal to

The solution of the differential equation $y\mathrm{d}x-\left(x+2y^2\right)\mathrm{d}y=0$ is $x=f\left(y\right)$. If $f\left(-1\right)=1,$ then $f\left(1\right)$ is equal to