Question

Let $y=y(x)$ satisfy the differential equation $(2xy+x^2y+\frac{y^3}{3})dx+(x^2+y^2)dy=0$. If $y(1)=1$ and the value of $(y(0))^3=ke (k \in N)$. Find $k$.

Answer 1

4

Answer 2

The given differential equation is $(2xy+x^2y+\frac{y^3}{3})dx+(x^2+y^2)dy=0$.

This is a first-order differential equation of the form $M(x,y)dx + N(x,y)dy = 0$, where $M(x,y) = 2xy+x^2y+\frac{y^3}{3}$ and $N(x,y) = x^2+y^2$.

First, we check if the equation is exact by comparing the partial derivatives: $\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(2xy+x^2y+\frac{y^3}{3}) = 2x+x^2+y^2$. $\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(x^2+y^2) = 2x$.

Since $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$, the equation is not exact.

Next, we look for an integrating factor. Consider the expression $\frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x})$: $\frac{1}{x^2+y^2}((2x+x^2+y^2) - 2x) = \frac{x^2+y^2}{x^2+y^2} = 1$. Since this expression is a function of $x$ only, an integrating factor $\mu(x)$ exists and is given by $e^{\int 1 dx} = e^x$.

Multiply the given differential equation by the integrating factor $e^x$: $e^x(2xy+x^2y+\frac{y^3}{3})dx+e^x(x^2+y^2)dy=0$. Let the new coefficients be $M'(x,y) = e^x(2xy+x^2y+\frac{y^3}{3})$ and $N'(x,y) = e^x(x^2+y^2)$. We verify that this new equation is exact: $\frac{\partial M'}{\partial y} = \frac{\partial}{\partial y}[e^x(2xy+x^2y+\frac{y^3}{3})] = e^x(2x+x^2+y^2)$. $\frac{\partial N'}{\partial x} = \frac{\partial}{\partial x}[e^x(x^2+y^2)] = e^x(x^2+y^2) + e^x(2x) = e^x(x^2+2x+y^2)$. Since $\frac{\partial M'}{\partial y} = \frac{\partial N'}{\partial x}$, the equation is exact.

Now we find the potential function $F(x,y)$ such that $dF = M'dx + N'dy$. We have $\frac{\partial F}{\partial y} = N'(x,y) = e^x(x^2+y^2)$. Integrate with respect to $y$ (treating $x$ as a constant): $F(x,y) = \int e^x(x^2+y^2) dy = e^x \int (x^2+y^2) dy = e^x (x^2y + \frac{y^3}{3}) + h(x)$, where $h(x)$ is an arbitrary function of $x$.

Now, differentiate $F(x,y)$ with respect to $x$ and set it equal to $M'(x,y)$: $\frac{\partial F}{\partial x} = \frac{\partial}{\partial x}[e^x (x^2y + \frac{y^3}{3}) + h(x)] = e^x (x^2y + \frac{y^3}{3}) + e^x (2xy) + h'(x)$. We are given $M'(x,y) = e^x(2xy+x^2y+\frac{y^3}{3})$. So, $e^x (x^2y + \frac{y^3}{3}) + e^x (2xy) + h'(x) = e^x(2xy+x^2y+\frac{y^3}{3})$. $e^x x^2y + e^x \frac{y^3}{3} + e^x 2xy + h'(x) = e^x 2xy + e^x x^2y + e^x \frac{y^3}{3}$. This simplifies to $h'(x) = 0$. Integrating with respect to $x$, we get $h(x) = C_0$, where $C_0$ is a constant.

The general solution is $F(x,y) = C_1$, where $C_1$ is a constant. $e^x (x^2y + \frac{y^3}{3}) + C_0 = C_1$. Let $C = C_1 - C_0$. The general solution is $e^x (x^2y + \frac{y^3}{3}) = C$.

We are given the initial condition $y(1)=1$. Substitute $x=1$ and $y=1$ into the general solution: $e^1 (1^2 \cdot 1 + \frac{1^3}{3}) = C$. $e (1 + \frac{1}{3}) = C$. $e (\frac{4}{3}) = C$. $C = \frac{4e}{3}$.

The particular solution is $e^x (x^2y + \frac{y^3}{3}) = \frac{4e}{3}$.

We need to find the value of $(y(0))^3$. Let $y_0 = y(0)$ be the value of $y$ when $x=0$. Substitute $x=0$ into the particular solution: $e^0 (0^2 \cdot y_0 + \frac{y_0^3}{3}) = \frac{4e}{3}$. $1 (0 + \frac{y_0^3}{3}) = \frac{4e}{3}$. $\frac{y_0^3}{3} = \frac{4e}{3}$. $y_0^3 = 4e$.

So, $(y(0))^3 = 4e$. We are given that $(y(0))^3 = ke$, where $k \in N$. Comparing $4e$ with $ke$, we have $ke = 4e$. Since $e \neq 0$, we can divide by $e$ to get $k=4$. The value $k=4$ is a natural number.