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Question: How do you find all solutions of the differential equation \(\dfrac{dy}{dx}=\dfrac{{{x}^{2}}}{y}?\)...

How do you find all solutions of the differential equation dydx=x2y?\dfrac{dy}{dx}=\dfrac{{{x}^{2}}}{y}?

Explanation

Solution

We use variables separable to find the solutions of the given equation. We take each variable and its derivative to the same side. After we separate the variables, we integrate the equation to get the solution.

Complete step-by-step solution:
Consider the given first order differential equation dydx=x2y.......(1)\dfrac{dy}{dx}=\dfrac{{{x}^{2}}}{y}.......\left( 1 \right)
Equation (1)\left( 1 \right) is called a separable equation, because the variables xx and yy can be separated so that xx appears only on one side (say right) and yy appears only on the other side (say left).
Let us separate the variables as follows,
⇒ydy=x2dx\Rightarrow ydy={{x}^{2}}dx
Now we have separated the variables into different sides.
Therefore, we are integrating the above obtained equation to get the solution.
⇒∫ydy=∫x2dx\Rightarrow \int{ydy=\int{{{x}^{2}}dx}}
Since ∫xndx=xn+1n+1+C\int{{{x}^{n}}d}x=\dfrac{{{x}^{n+1}}}{n+1}+C where CC is the constant of integration, we get
⇒y22=x33+C\Rightarrow \dfrac{{{y}^{2}}}{2}=\dfrac{{{x}^{3}}}{3}+C
Now we got this, we are transposing 22 from the left side to the right side,
⇒y2=2x33+C1\Rightarrow {{y}^{2}}=\dfrac{2{{x}^{3}}}{3}+{{C}_{1}} where C1=2C{{C}_{1}}=2C
Take square root of the obtained equation,
⇒y=2x23+C1\Rightarrow y=\sqrt{\dfrac{2{{x}^{2}}}{3}+{{C}_{1}}}
Hence, the solution of the given first order differential equation is y=2x23+C1.y=\sqrt{\dfrac{2{{x}^{2}}}{3}+{{C}_{1}}}.

Note: The first order differential equation (1)\left( 1 \right) is called a separable equation or an equation with separable variables. More generally, a first order differential equation of the form dydx=M(x)N(y)\dfrac{dy}{dx}=\dfrac{M\left( x \right)}{N\left( y \right)} is called a separable equation or an equation with separable variables. We solve it this way:
⇒dydx=M(x)N(y).\Rightarrow \dfrac{dy}{dx}=\dfrac{M\left( x \right)}{N\left( y \right)}.
By algebraic manipulation (we can simply say, separation of variables) we get,
⇒N(y)dy=M(x)dx.\Rightarrow N\left( y \right)dy=M\left( x \right)dx.
We integrate this equation to get the solution.
⇒∫N(y)dy=∫M(x)dx+C\Rightarrow \int{N\left( y \right)dy=\int{M\left( x \right)dx+C}}, where CC is the constant of integration for the indefinite integrals.
Remember that many first order differential equations can be reduced to form the separable equation dydx=M(x)N(y).\dfrac{dy}{dx}=\dfrac{M\left( x \right)}{N\left( y \right)}.