Video: Finding the Particular Solution of a Differential Equation Given a Point the Solution Passes Through

Find the solution of the differential equation (12 d𝑦/dπ‘₯) + (𝑦²(𝑒^(π‘₯))) = 0 that passes through the point (0, 4).

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Video Transcript

Find the solution of the differential equation 12 d𝑦 by dπ‘₯ plus 𝑦 squared 𝑒 to the power of π‘₯ is equal to zero that passes through the point zero, four.

The question wants us to find the specific solution to the differential equation 12 d𝑦 by dπ‘₯ plus 𝑦 squared times 𝑒 to the power of π‘₯ is equal to zero, which passes through the point zero, four. By subtracting 𝑦 squared times 𝑒 to the power of π‘₯ from both sides of our equation, we see that 12 d𝑦 by dπ‘₯ is equal to negative 𝑦 squared 𝑒 to the power of π‘₯. We can then divide through by 12, and we see that this is a separable differential equation.

And we recall we call a differential equation separable if its first order, and we can write it in the form d𝑦 by dπ‘₯, is equal to a product of two functions, one in π‘₯ and one in 𝑦. To solve these types of differential equations, we want to separate the 𝑦-variable and the π‘₯-variable onto opposite sides of the equation.

So we’ll start by dividing both sides of our equation through by our function in 𝑦; that’s 𝑦 squared. This gives us one over 𝑦 squared d𝑦 by dπ‘₯ is equal to negative 𝑒 to the power of π‘₯ divided by 12. And it’s worth reiterating at this point that d𝑦 by dπ‘₯ is not a fraction. However, when we’re solving separable differential equations, we can treat it a little bit like a fraction. This gives us the equivalent statement, one over 𝑦 squared d𝑦 is equal to negative 𝑒 to the power of π‘₯ over 12 dπ‘₯. To solve this, we want to integrate both sides of our equation. And we see we can evaluate both of these integrals.

Using our laws of exponents, we know one over 𝑦 squared is the same as saying 𝑦 to the power of negative two. We can then evaluate this integral by using our power rule for integration. We add one to the exponent and then divide by this new exponent. Adding one to our exponent of negative two gives us negative one. And then we divide by this value of negative one. Finally, we add a constant of integration we will call 𝑐 one. And we know the integral of 𝑒 to the power of π‘₯ is just equal to itself. So the integral of negative 𝑒 to the power of π‘₯ over 12 is just equal to negative 𝑒 to the power of π‘₯ over 12 plus a constant of integration we will call 𝑐 two.

We can simplify this. We know dividing by negative one is the same as multiplying by negative one. And we can combine the constants of integration, 𝑐 one and 𝑐 two, into a new constant we will call 𝑐. This gives us negative 𝑦 to the power of negative one is equal to negative 𝑒 to the power of π‘₯ over 12 plus 𝑐. Now, we remember the question wanted us to find the specific solution to our differential equation which passes through the point zero, four. This is the same as saying when π‘₯ is equal to zero, 𝑦 is equal to four. We can use this to find the value of our constant 𝑐.

We substitute π‘₯ is equal to zero and 𝑦 is equal to four into our general solution for the differential equation. This gives us negative four to the power of negative one minus 𝑒 to zeroth power over 12 plus 𝑐. We know 𝑒 to the zeroth power is equal to one and four to the power of negative one is one-quarter. So we can rearrange this equation to get negative one-quarter plus one twelfth is equal to 𝑐. And we can then calculate this to be negative one-sixth. So by substituting 𝑐 is equal to negative one-sixth into our general solution of the differential equation, we get the specific solution, negative 𝑦 to the power of negative one is equal to negative 𝑒 to the power of π‘₯ over 12 minus one-sixth.

We could leave our answer like this. However, we can rearrange it to give 𝑦 in terms of π‘₯. We multiply both sides of our equation through by negative one. This gives us the reciprocal of 𝑦 is equal to 𝑒 to the power of π‘₯ over 12 plus one-sixth. We rewrite one-sixth; that’s two divided by 12. So we can add our two fractions together. This gives us the reciprocal of 𝑦 is equal to 𝑒 to the power of π‘₯ plus two over 12.

Finally, we take the reciprocal of both sides of the equation to see that the specific solution to the differential equation 12 d𝑦 by dπ‘₯ plus 𝑦 squared 𝑒 to the power of π‘₯ is equal to zero, which passes through the point zero, four, is given by 𝑦 is equal to 12 divided by 𝑒 to the power of π‘₯ plus two.

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