Methods of Solving First Order, First Degree Differential Equation

Let $I$ be the purchase value of equipment and $V\left(t\right)$ be the value after it has been used for $t \mathrm{years}$. The value $V\left(t\right)$ depreciates at a rate given by the differential equation

$\frac{dV\left(t\right)}{dt}=-k\left(T-t\right)$

Where, $k > 0$ is a constant and $T$ is the total life in years of the equipment. Then the scrap value $V\left(T\right)$ of the equipment is:

Let $f:[1.\infty )\to \mathrm{ℝ}$ be a differentiable function such that $f\left(1\right)=\frac{1}{3}$ and $3{\int }_1^{x}f\left(t\right)dt=xf\left(x\right)-\frac{x^3}{3},x\in [1,\infty )$. Let $e$ denote the base of the natural logarithm. Then the value of $f\left(e\right)$ is

A solution of the differential equation ${\left(\frac{dy}{dx}\right)}^2-x\frac{dy}{dx}=0$ is

If $y=y\left(x\right)$ is the solution of the differential equation $\frac{dy}{dx}+\frac{4x}{\left(x^2-1\right)}y=\frac{x+2}{{\left(x^2-1\right)}^{\frac{5}{2}}},x>1$ such that $y\left(2\right)=\frac{2}{9}{\log }_e\left(2+\sqrt{3}\right)$ and $y\left(\sqrt{2}\right)=\alpha {\log }_e\left(\sqrt{\alpha }+\beta \right)+\beta -\sqrt{\gamma },\alpha ,\beta ,\gamma \in \mathrm{\mathbb{N}}$, then $\alpha \beta \gamma$ is equal to

Let the tangent at any point $\mathrm{P}$ on a curve passing through the points $\left(1,1\right)$ and $\left(\frac{1}{10},100\right)$, intersect positive $\mathrm{x}$-axis and $\mathrm{y}$-axis at the points $\mathrm{A}$ and $\mathrm{B}$ respectively. If $P A: P B=1: k$ and $\mathrm{y}=\mathrm{y}\left(\mathrm{x}\right)$ is the solution of the differential equation $e^{\frac{dy}{dx}}=kx+\frac{k}{2},y\left(0\right)=k$, then $4\mathrm{y}\left(1\right)-5{\log }_{\mathrm{e}}3$ is equal to _______________

Let$y=y\left(x\right)$ be the solution of the differential equation $\frac{dy}{dx}+\frac{5}{x\left(x^5+1\right)}y=\frac{{\left(x^5+1\right)}^2}{x^7}, x>0$. If $y\left(1\right)=2$, then $y\left(2\right)$ is equal to

Let $y=y_1\left(x\right)$ and $y=y_2\left(x\right)$ be the solution curves the differential equation $\frac{dy}{dx}=y+7$ with initial conditions $y_1\left(0\right)=0$ and $y_2\left(0\right)=1$ respectively. Then the curves $y=y_1\left(x\right)$ and $y = y_2\left(x\right)$ intersect at

Let $y=y\left(x\right)$ be a solution curve of the differential equation, $\left(1-x^2{y}^2\right)dx=ydx+xdy$, If the line $x=1$ intersects the curve $y=y\left(x\right)$ at $y=2$ and the line $x=2$ intersects the curve $y=y\left(x\right)$ at $y=\alpha$, then a value of $\alpha$ is

Let $x=x\left(y\right)$ be the solution of the differential equation $2\left(y+2\right){\log }_e\left(y+2\right)dx+\left(x+4-2{\log }_e\left(y+2\right)\right)dy=0$, $y>-1$ with $x\left(e^4-2\right)=1$. Then $x\left(e^9-2\right)$ is equal to

The slope of tangent at any point $\left(x, y\right)$ on a curve $y=y\left(x\right)$ is $\frac{x^2+y^2}{2xy}, x>0$. If $y\left(2\right)=0$, then a value of $y\left(8\right)$ is

Let a curve $y=f\left(x\right), x\in \left(0, \infty \right)$ pass through the points $P\left(1, \frac{3}{2}\right)$ and $Q\left(a, \frac{1}{2}\right)$. If the tangent at any point $R(b, f(b\left)\right)$ to the given curve cuts the $y$-axis at the point $S(0, c)$ such that $bc=3,$ then ${\left(PQ\right)}^2$ is equal to _____.

Solve the differential equation $\frac{dy}{dx}-\frac{3y}{x}+\frac{6}{x}=0$.

Solve the following differential equation $\frac{dy}{dx}=\frac{1+x^2}{1+y^2}$

If $\frac{dy}{dx}=6e^x+e^{2x}+e^{3x}$, then $y\left(2\right)-y\left(0\right)$ is

If $\frac{dy}{dx}=y+7$ and $y\left(0\right)=0$, then the value of $y\left(1\right)$ is

For $\frac{dy}{dx}+\frac{5}{x\left(1+x^5\right)}y=\frac{{\displaystyle {\left(1+x^5\right)}^2}}{x^7}$; $y\left(1\right)=2$, then the value of $y\left(2\right)$ is

Slope of tangent to a curve at a variable point is $\frac{x^2+y^2}{2xy}$ and $y\left(2\right)=0$, then $y\left(8\right)$ is

Let $y=y\left(x\right)$ be the solution of the differential equation $\left(3y^2-5x^2\right)ydx+2x\left(x^2-y^2\right)dy=0$ such that $y\left(1\right)=1$, then $\left|{\left(y\left(2\right)\right)}^3-12y\left(2\right)\right|$ is equal to:

Let $y=y\left(x\right)$ be the solution of the differential equation $\frac{dy}{dx}=\frac{4y^3+2y{x}^2}{3x{y}^2+x^3}; y\left(1\right)=1$. If for some $n\in N$, $y\left(2\right)\in [n-1,n)$

For what value of $n$ is the $\frac{dy}{dx}=\frac{x^3-y^n}{x^{2}y+x{y}^2}$ a homogeneous differential equation: