Question

How do you find the general solution of $\dfrac{dy}{dx}={{x}^{3}}-4x$ ?

Answer 1

Hint: We have to find the general solution for $\dfrac{dy}{dx}={{x}^{3}}-4x$ . We can solve this by variable separation we can keep the x variable in LHS and y in RHS and integrate both sides to find the solution

Complete step by step answer:
The given equation in the question is $\dfrac{dy}{dx}={{x}^{3}}-4x$
Multiplying dx in both LHS and RHS we get
$dy=\left( {{x}^{3}}-4x \right)dx$
Now we can integrate both sides
If f and g are 2 different function then we know that the value of $\int{\left( f\left( x \right)\pm g\left( x \right) \right)}dx$ is equal to the value of $\int{f\left( x \right)dx\pm \int{g\left( x \right)dx}}$ and if a constant is multiplied to the function we can take the constant outside the integration.
So we can write $\left( {{x}^{3}}-4x \right)dx$ as $\int{{{x}^{3}}dx-\int{4xdx}}$ and also we can write $\int{4xdx=4\int{xdx}}$
We know that integration of ${{x}^{n}}dx$ is equal to $\dfrac{{{x}^{n+1}}}{n+1}$ where n is a real number and n is not equal to -1.
Applying the above formula we can tell that the value integration of ${{x}^{3}}$ is $\dfrac{{{x}^{4}}}{4}$ and integration of x is $\dfrac{{{x}^{2}}}{2}$
We are solving a indefinite integration here so we have to add a constant here
Now we can write $\int{dy}=\int{\left( {{x}^{3}}-4x \right)dx}$
$\Rightarrow ~y=\int{{{x}^{3}}dx}-\int{4xdx+c}$
$\Rightarrow y=\dfrac{{{x}^{4}}}{4}-2{{x}^{2}}+c$
So the general solution of the differential equation $\dfrac{dy}{dx}={{x}^{3}}-4x$ is $y=\dfrac{{{x}^{4}}}{4}-2{{x}^{2}}+c$

Note:
While writing the integration of ${{x}^{n}}$ is equal to $\dfrac{{{x}^{n+1}}}{n+1}$ keep in mind that n is not equal to -1
If we put -1 in the formula $\dfrac{{{x}^{n+1}}}{n+1}$ the denominator will be equal to 0 and that will be not defined. The integration of ${{x}^{n}}$ when n is equal to -1 is ln x, derivative of ln x is $\dfrac{1}{x}$.