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- 1. Differential EquationsDr.G.V.RAMANA REDDYM.Sc, Ph.DKL University,GunturA.P
- 2. Differential Equations( ) ( )4 2 2 3sin , 2 0, 0y x y y xy x y y x′ ′′= − + − = + + =Definition A differential equation is an equation involvingderivatives of an unknown function and possibly thefunction itself as well as the independent variable.ExampleDefinition The order of a differential equation is the highest orderof the derivatives of the unknown functionappearing in the equation1storder equations 2ndorder equation( ) ( )′ = ⇒ = − +sin cosy x y x CExamples′′ ′= + ⇒ = + + ⇒ = + + +2 31 1 26 e 3 e ex x xy x y x C y x C x CIn the simplest cases, equations may be solved by direct integration.Observe that the set of solutions to the above 1storder equation has 1parameter, while the solutions to the above 2ndorder equationdepend on two parameters.
- 3. Separable Differential EquationsA separable differential equation can be expressed asthe product of a function of x and a function of y.( ) ( )dyg x h ydx= ×Example:22dyxydx=Multiply both sides by dx and divideboth sides by y2to separate thevariables. (Assume y2is never zero.)22dyx dxy=22y dy x dx−=( ) 0h y ≠
- 4. A separable differential equation can be expressed asthe product of a function of x and a function of y.( ) ( )dyg x h ydx= ×Example:22dyxydx=22dyx dxy=22y dy x dx−=22y dy x dx−=∫ ∫1 21 2y C x C−− + = +21x Cy− = +21yx C− =+ 21yx C= −+( ) 0h y ≠→CombinedconstantsofintegrationSeparable Differential Equations
- 5. Family of solutions (general solution)of a differential equationExampleThe picture on the right shows somesolutions to the above differentialequation. The straight linesy = x and y = -xare special solutions. A uniquesolution curve goes through anypoint of the plane different fromthe origin. The special solutions y= x and y = -x go both throughthe origin.Cxyxdxydyyxdxdy+=== ∫ ∫22
- 6. Initial conditions• In many physical problems we need to find the particularsolution that satisfies a condition of the form y(x0)=y0.This is called an initial condition, and the problem offinding a solution of the differential equation thatsatisfies the initial condition is called an initial-valueproblem.• Example (cont.): Find a solution to y2= x2+ C satisfyingthe initial condition y(0) = 2.22= 02+ CC = 4y2= x2+ 4
- 7. Example:( )222 1 xdyx y edx= +22121xdy x e dxy=+Separable differential equation22121xdy x e dxy=+∫ ∫2u x=2du x dx=211udy e duy=+∫ ∫11 2tan uy C e C−+ = +211 2tan xy C e C−+ = +21tan xy e C−= + Combined constants of integration →
- 8. Example (cont.):( )222 1 xdyx y edx= +21tan xy e C−= + We now have y as an implicitfunction of x.We can find y as an explicit functionof x by taking the tangent of bothsides.( ) ( )21tan tan tan xy e C−= +( )2tan xy e C= +→
- 9. A population of living creatures normally increases at arate that is proportional to the current level of thepopulation. Other things that increase or decrease at arate proportional to the amount present includeradioactive material and money in an interest-bearingaccount.If the rate of change is proportional to the amount present,the change can be modeled by:dykydt=→Law of natural growth or decay
- 10. dykydt=1dy k dty=1dy k dty=∫ ∫ln y kt C= +Rate of change is proportionalto the amount present.Divide both sides by y.Integrate both sides.→ln y kt Ce e += Exponentiate both sides.C kty e e= ⋅C kty e e= ±kty Ae=
- 11. Real-life populations do not increase forever. There issome limiting factor such as food or living space.There is a maximum population, or carrying capacity, M.A more realistic model is the logistic growth model wheregrowth rate is proportional to both the size of thepopulation (y) and the amount by which y falls short ofthe maximal size (M-y). Then we have the equation:)( yMkydtdy−=Logistic Growth ModelThe solution to this differential equation (derived in the textbook):)0(where,)(0000yyeyMyMyy kMt=−+= −
- 12. A tank contains 1000 L of brine with 15 kg of dissolvedsalt. Pure water enters the tank at a rate of 10 L/min.The solution is kept thoroughly mixed and drains fromthe tank at the same rate.How much salt is in the tank(a) after t minutes;(b) after 20 minutes?This problem can be solved by modeling it as a differentialequation.(the solution on the board)Mixing Problems
- 13. Problem 45.A vat with 500 gallons of beer contains 4%alcohol (by volume). Beer with 6%alcohol is pumped into the vat at a rate of5 gal/min and the mixture is pumped outat the same rate. What is the percentageof alcohol after an hour?Mixing Problems

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