The first thing we must do is rewrite the equation:

$\displaystyle \left(\frac{1}{y^2}\right)dy=(6x^2)dx$

We can then find the integrals:

$\displaystyle \int \left(\frac{1}{y^2}\right)dy=\int (6x^2)dx$

  The integrals are as follows:

$\displaystyle \int \left(\frac{1}{y^2}\right)dy=-\frac{1}{y}$

$\displaystyle \int (6x^2)dx=2x^3$

We're left with:

$\displaystyle -\frac{1}{y}=2x^3+c$

We then plug in the initial condition and solve for $\displaystyle c.$

$\displaystyle -\frac{1}{(-\frac{1}{4})}=2\left(-\frac{1}{2}\right)^3+c$

$\displaystyle 4=-\frac{1}{4}+c$

$\displaystyle c=\frac{17}{4}$

The particular solution is then:

$\displaystyle -\frac{1}{y}=2x^3+\frac{17}{4}$

Remember: $\displaystyle \sqrt{xy}=(\sqrt{x} ) (\sqrt{y})$

The first thing we must do is rewrite the equation:

$\displaystyle \left(\frac{1}{\sqrt{y}}\right)dy=(\sqrt{x})dx$

We can then find the integrals:

$\displaystyle \int \left(\frac{1}{\sqrt{y}}\right)dy=\int (\sqrt{x})dx$

The integrals are as follows:

$\displaystyle \int \left(\frac{1}{\sqrt{y}}\right)dy= \int y^{-1/2}dy=\frac{y^{1/2}}{\frac{1}{2}}=2y^{1/2}=2\sqrt{y}$

$\displaystyle \int (\sqrt{x})dx=\frac{2}{3}(x^{\frac{3}{2}})$

We're left with:

$\displaystyle 2\sqrt{y}=\frac{2}{3}(x)^{\frac{3}{2}}+c$

We then plug in the initial condition and solve for $\displaystyle c.$

$\displaystyle 2\sqrt{\frac{256}{9}}= \frac{2}{3}(4)^{\frac{3}{2}}+c$

$\displaystyle \frac{32}{3}=\frac{16}{3}+c$

$\displaystyle c=\frac{16}{3}$

The particular solution is then:

$\displaystyle 2\sqrt{y}=\frac{2}{3}(x)^{\frac{3}{2}}+\frac{16}{3}$

The first thing we must do is rewrite the equation:

$\displaystyle (y^2+4y^3)dy=(5x^4)dx$

We can then find the integrals:

$\displaystyle \int (y^2+4y^3)dy=\int (5x^4)dx$

The integrals are as follows:

$\displaystyle \int (y^2+4y^3)dy=\frac{1}{3}y^3+y^4$

$\displaystyle \int (5x^4)dx=x^5$

We're left with

$\displaystyle \frac{1}{3}y^3+y^4=x^5+c$

We plug in the initial condition and solve for $\displaystyle c.$

$\displaystyle \frac{1}{3}(0)^3+(0)^4=(1)^5+c$

$\displaystyle c=-1$

The particular solution is then:

$\displaystyle \frac{1}{3}y^3+y^4=x^5-1$

The first thing we must do is rewrite the equation:

$\displaystyle ({y^5+3y^2-2y})dy=({2x^3+3x^2-x+7})dx$

We can then find the integrals:

$\displaystyle ({y^5+3y^2-2y})dy=({2x^3+3x^2-x+7})dx$

The integrals are as follows:

$\displaystyle \int ({y^5+3y^2-2y})dy=\frac{1}{6}y^6+y^3-y^2$

$\displaystyle \int ({2x^3+3x^2-x+7})dx=\frac{1}{2}x^4+x^3-\frac{1}{2}x^2+7x$

We're left with

$\displaystyle \frac{1}{6}y^6+y^3-y^2=\frac{1}{2}x^4+x^3-\frac{1}{2}x^2+7x+c$

Plugging in the initial conditions and solving for c gives us:

$\displaystyle \frac{1}{6}+1-1=8-8-2+14+c$

$\displaystyle c=-\frac{167}{6}$

The particular solution is then,

$\displaystyle \frac{1}{6}y^6+y^3-y^2=\frac{1}{2}x^4+x^3-\frac{1}{2}x^2+7x-\frac{167}{6}$

Using the power rule, we can differentiate our first term reducing the power by one and multiplying our term by the original power. $\displaystyle x^{2}$, will thus become $\displaystyle 2x$. The second term $\displaystyle 4x$, will thus become $\displaystyle 4$. The last term is a constant value, so according to the power rule this term will become $\displaystyle 0$.

We will use the fact that $\displaystyle e^{au} = ae^{au}$ to differentiate. Let $\displaystyle a=6$ and $\displaystyle u=x$. Substituing our values we can see the derivative will be $\displaystyle 6e^{6x}$.

Using the product rule, we determine the derivative of $\displaystyle f(x)\cdot g(x) = {f}'(x)\cdot g(x) + f(x)\cdot {g}'(x))$
Let $\displaystyle f(x) = x^{2}$ and $\displaystyle g(x) = \sin(x)$. We can see that $\displaystyle f{}'(x) = 2x$ and $\displaystyle g{}'(x) = \cos(x)$.

Plugging in our values into the product rule formula, we are left with the final derivative of $\displaystyle x (2 \sin(x)+x \cos(x))$.

According to the power rule, whenever we differentiate a constant value it will reduce to zero. Since the only term of our function is a constant, we can only differentiate  $\displaystyle 74 \rightarrow 0$.

Using the chain rule, we will differentiate the exponent of our exponential function, and then multiply our original function. Differentiating our exponent with the power rule will yield $\displaystyle 4x$. Using the chain rule we will multiply this by our original function resulting in $\displaystyle 4xe^{2x^{2}}$.

Using the power rule, we can differentiate our first term reducing the power by one and multiplying our term by the original power. $\displaystyle x^{2}$, will thus become $\displaystyle 2x$. The second term is a constant value, so according to the power rule this term will become $\displaystyle 0$.