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Question: How do you solve the differential equation \(\dfrac{{dy}}{{dx}} = 6{y^2}x,\) where \(y(1) = \dfrac{1...

How do you solve the differential equation dydx=6y2x,\dfrac{{dy}}{{dx}} = 6{y^2}x, where y(1)=125y(1) = \dfrac{1}{{25}}?

Explanation

Solution

Here first of all we will separate both the variables on different sides and then will apply integration on both the sides and will simplify the equation and place the given coordinates to find the constant value and then replace in the final solution.

Complete step-by-step solution:
Take the given expression: dydx=6y2x\dfrac{{dy}}{{dx}} = 6{y^2}x
Now separate both the variables “x” and “y” from the above equation. Remember when you move the term multiplicative on one side to the opposite side then it goes to the denominator and similarly term in division on one side if moved to the opposite side then it goes to the numerator.
⇒1y2dy=6xdx\Rightarrow \dfrac{1}{{{y^2}}}dy = 6xdx
Now take integration on both the sides of the equation.
∫1y2dy=∫6xdx\int {\dfrac{1}{{{y^2}}}dy} = \int {6xdx}
Apply the formula, ∫xndx=xn+1n+1\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} in the above equation
⇒−1y=3x2+c\Rightarrow - \dfrac{1}{y} = 3{x^2} + c
Multiply both the sides of the above equation with minus one
⇒1y=−(3x2+c)\Rightarrow \dfrac{1}{y} = - (3{x^2} + c)
Cross multiply and make “y” the subject, in cross multiplication denominator of one side is multiplied with the numerator of the opposite side.
⇒y=−1(3x2+c)\Rightarrow y = - \dfrac{1}{{(3{x^2} + c)}} ….. (A)
We are also given that y(1)=125y(1) = \dfrac{1}{{25}}so place x=1x = 1in the above equation.
⇒125=−1(3(1)2+c)\Rightarrow \dfrac{1}{{25}} = - \dfrac{1}{{(3{{(1)}^2} + c)}}
Simplify the above equation, multiply with minus one on both the sides of the equation.
⇒−125=13+c\Rightarrow - \dfrac{1}{{25}} = \dfrac{1}{{3 + c}}
Do cross multiplication
⇒3+c=−25\Rightarrow 3 + c = - 25
Make “c” the subject and move on the opposite side. Move the constant term on the left hand side of the equation from the right hand side of the equation. When you move any term from one side to another then the sign of the term also changes. Positive terms become negative and the negative term becomes positive.
⇒c=−25−3\Rightarrow c = - 25 - 3
While simplifying two negative numbers in actual you have to add both the numbers and give a negative sign to the resultant answer.
⇒c=−28\Rightarrow c = - 28
Place the above value in the equation (A)
⇒y=−1(3x2−28)\Rightarrow y = - \dfrac{1}{{(3{x^2} - 28)}}
Applying the negative sign inside the bracket which changes the sign of all the terms
⇒y=128−3x2\Rightarrow y = \dfrac{1}{{28 - 3{x^2}}}
This is the required solution.

Note: Know the difference between the differentiation and the integration and apply the formula accordingly. integration is the concept of calculus and it is the act of finding the integrals whereas Differentiation can be represented as the rate of change of the function, whereas integration represents the sum of the function over the range. They are inverses of each other. Be careful about plus and minus signs.