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Question

Question: What is a solution to the differential equation \(\dfrac{{dy}}{{dx}} = {x^2}y\)?...

What is a solution to the differential equation dydx=x2y\dfrac{{dy}}{{dx}} = {x^2}y?

Explanation

Solution

Hint : In the given question, we are given a differential equation. So, we have to solve the given differential equation using methods of integration. We will classify the differential equation and then solve according to the type of equation. Then, we will add an arbitrary constant c which is put after computing an indefinite integral as it represents a family of curves.

Complete step-by-step answer :
The given question requires us to solve a differential equation dydx=x2y\dfrac{{dy}}{{dx}} = {x^2}y using methods of integration. First of all, we have to classify the type of differential equation.
Now, we observe that we can separate the variables in the differential equation given to us. So, the differential equation dydx=x2y\dfrac{{dy}}{{dx}} = {x^2}y is of variable separable form.
Now, we separate the variables. So, we get,
dyy=dx(x2)\Rightarrow \int {\dfrac{{dy}}{y}} = \int {dx\left( {{x^2}} \right)}
Now, integrating both sides using the power rule of integration xndx=xn+1n+1\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}, we get,
dyy=x33+c\Rightarrow \int {\dfrac{{dy}}{y}} = \dfrac{{{x^3}}}{3} + c
We also know that the integral of logarithmic function logx\log x with respect to x is 1x\dfrac{1}{x}. So, we get,
lny=x33+c\Rightarrow \ln y = \dfrac{{{x^3}}}{3} + c, where c is any arbitrary constant.
Therefore, a solution of the differential equation dydx=x2y\dfrac{{dy}}{{dx}} = {x^2}y is lny=x33+c\ln y = \dfrac{{{x^3}}}{3} + c
So, the correct answer is “lny=x33+c\ln y = \dfrac{{{x^3}}}{3} + c”.

Note : The indefinite integrals of certain functions may have more than one answer in different forms. However, all these forms are correct and interchangeable into one another. Indefinite integral gives us the family of curves as we don’t know the exact value of the constant. Exact value of the arbitrary constant can be calculated only if we are given a point lying on the function.