# Question Video: Solving a Separable First Order Differential Equations given Its Initial Value Mathematics • Higher Education

Find the solution of the differential equation dπ/dπ‘ = β(ππ‘) that satisfies the initial condition π(1) = 2.

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### Video Transcript

Find the solution to the differential equation dπ by dπ‘ is equal to the square root of ππ‘ that satisfies the initial condition π of one is equal to two.

The question wants us to find a specific solution to the differential equation dπ by dπ‘ is equal to the square root of π times π‘ using the initial condition that π of one is equal to two. We can see that our differential equation is almost already the product of a function in π and the product of a function in π‘. In fact, by using our laws of exponents, we can see this more clearly. dπ by dπ‘ is equal to the square root of π multiplied by the square root of π‘.

Since this is a first-order differential equation which can be written as the products of a function in π and a function in π‘, we call this a separable differential equation. To solve this, we want to separate our variables π and π‘ onto opposite sides of the equation. Weβll start by dividing through by the square root of π. This gives us that one divided by the square root of π times dπ by dπ‘ is equal to the square root of π‘. Although dπ by dπ‘ is not a fraction, when weβre solving separable differential equations, we can treat it a little bit like a fraction.

Using this, we get the equivalent statement one divided by the square root of π dπ is equal to the square root of π‘ dπ‘. We then want to integrate both sides of this equation. We could then evaluate both of these integrals by using the power rule for integration, which tells us if π is not equal to negative one, the integral of π₯ to the πth power with respect to π₯ is equal to π₯ to the power of π plus one divided by π plus one plus a constant of integration π. We add one to the exponent and then divide by this new exponent.

To see why we can do this, we know that one divided by the square root of π is the same as saying π to the power of negative one-half and the square root of π‘ is the same as saying π‘ to the power of one-half. To integrate π to the power of negative a half with respect to π, we add one to our exponent and then divide by this new exponent and add a constant of integration we will call π one. And we can simplify this since negative a half plus one is just equal to one-half. And we can simplify this further since dividing by a fraction is the same as multiplying by the reciprocal. So, dividing by one-half is the same as multiplying by two.

This gives us two π to the power of a half plus π one. We then do the same to integrate π‘ to the power of a half with respect to π‘. We add one to our exponent and then divide by this new exponent. And we add a constant of integration we will call π two. We can simplify this expression further. We can combine the constants π one and π two into a new constant we will call π. We can simplify this further since one-half plus one is just equal to three over two.

Finally, just as we did before, instead of dividing by three over two, we can multiply by two divided by three. So, we now have the equation two π to the power of a half is equal to two-thirds π‘ to the power of three over two plus π. And we can actually find the value of π since the question gives us the initial condition π of one is equal to two. π is a function of π‘, so this is actually telling us when π‘ is equal to one, π is equal to two.

Substituting π‘ is equal to one and π is equal to two into our equation gives us two times two to the power of one-half is equal to two-thirds one to the power of three over two plus π. We can simplify this expression slightly. Two to the power of one-half is the same as saying the square root of two. And one to the power of any number is just equal to one. So, one to the power of three over two is one. So, this gives us that two root two is equal to two-thirds plus π. We can then subtract two-thirds from both sides of this equation to see that π is equal to negative two-thirds plus two root two.

Substituting this value of π into our equation gives us that two root π is equal to two-thirds π‘ to the power of three over two minus two-thirds plus two root two. And we could leave our answer like this. However, we see that every term shares a factor of two. So, we could just divide through by two. Dividing both sides of this equation through by two gives us the solution to the differential equation dπ by dπ‘ is equal to the square root of ππ‘, which satisfies the initial condition π of one is equal to two. As the square root of π is equal to one-third π‘ to the power of three over two minus one-third plus the square root of two.