 Hello and welcome to the session. In this session we will discuss equations reducible to homogeneous equations. Now we know that a differential equation of the form dy over dx is equal to f of xy over g of xy where f of xy and g of xy are homogeneous functions xy and of same degree. So the differential equation of the form dy over dx is equal to f of xy over g of xy is actually a homogeneous equation. For example dy over dx is equal to x square plus y square whole upon xy is a homogeneous equation as x square plus y square as well as xy are homogeneous functions of same degree. So the degree of x square plus y square is 2 and the degree of xy is also 2. That's why this equation is a homogeneous equation. Now let us discuss how to reduce the equations to the homogeneous form. Now the equation of the form dy over dx is equal to ax plus vy plus c whole upon capital A into x plus capital B into y plus capital C can be reduced to homogeneous form by changing the variables xy small x is equal to capital X plus h and small y is equal to capital Y plus k where h and k are the constants which are to be chosen so as to make the given equation homogeneous. Now dy over dx will be equal to now putting the value of y it will be the derivative of capital Y plus k with respect to x which is further equal to the derivative of capital Y with respect to x which can be further written as derivative of capital Y with respect to capital X into derivative of capital X with respect to x which is further equal to derivative of capital Y with respect to capital X because derivative of capital X with respect to x is equal to 1. Now putting the values of xy and dy over dx in this equation let it be equation number 1 we get the derivative of capital Y with respect to capital X is equal to A into capital X plus B into capital Y plus A h plus B k plus C the whole form capital A into x that is capital A into capital X plus capital B into capital Y plus capital A h plus capital B k plus capital C the whole now let h and k be chosen so as to satisfy the equations A h plus B k plus C is equal to 0 and capital A h plus capital B k plus capital C is equal to 0. Now solving x and y we get h is equal to B into capital C minus capital B into C over A into capital B minus capital A into B and k is equal to C into capital A minus A into capital C whole upon A into capital B minus capital A into B. Now this always has a meaning except when this is equal to 0 that is when A over capital A is equal to B over capital B. Now for this expression equal to 0 the equation will be reduced to a homogeneous equation that is the homogeneous equation will be the derivative of capital Y so respect to capital X is equal to A into capital X plus B into capital Y whole upon capital A into capital X plus capital B into capital Y. Now using the given equation to a homogeneous equation that is of this type we can easily solve this equation that is P by substituting capital Y is equal to V into capital X where V is any variable. Now let us discuss this procedure with the help of an example. Now here we have to solve the differential equation of the type V by over dx is equal to x minus 3y plus 4 whole upon 3x plus y minus 7. Now we will reduce this equation to the homogeneous form. Now the equation which is given in the example is of this form. Now first of all we will check that A over capital A is equal to B over capital B or not. So here A over capital A which is 1 by 3 is not equal to B over capital B that is minus 3 by 1 so by this procedure which we have discussed earlier we can solve this example. Now we will start with the solution. Now let this be equation Q supporting X is equal to capital X plus H and Y is equal to capital Y plus K is U. Now by using Y is equal to capital Y plus K we will get dy by dx is equal to the derivative of capital Y with respect to capital X. So the derivative of capital Y with respect to capital X will be equal to capital X plus H minus 3 into capital Y plus K the whole plus 4 whole upon 3 into capital X plus H the whole plus capital Y plus K minus 7. Now this implies the derivative of capital Y with respect to capital X is equal to capital X minus 3 into capital Y plus H minus 3K plus 4 whole upon 3 into capital X plus capital Y plus 3H plus K minus 7. Now to find H and K let H minus 3K plus 4 is equal to 0 that is this expression is equal to 0 and also 3H plus K minus 7 is equal to 0. Now solving 1 and 2 by the method of cross multiplication we get H over 21 minus 4 is equal to K over 12 minus of minus 7 that is 12 plus 7 is equal to 1 over 1 minus of minus 3 into 3 that is plus 9. On solving this implies H is equal to 17 by 10 and K is equal to 19 by 10. Now let this be equation R therefore for the values of H and K the derivative of capital Y with respect to capital X that is in equation R it will be equal to capital X minus 3 into capital Y whole upon 3 into capital X plus capital Y. Now putting capital Y is equal to V into capital X in this equation that is S VH. Now since we have taken capital Y is equal to V into capital X therefore the derivative of capital Y with respect to capital X is equal to V plus capital X into the the derivative of V with respect to capital X therefore we have V plus into the derivative of V with respect to capital X is equal to capital X minus 3 into V into capital X the whole upon 3 into capital X plus V into capital X which further implies capital X into the derivative of V with respect to capital X is equal to now taking X common and solving it will be 1 minus 3 V whole upon 3 plus V minus V and further on solving this implies capital X into the derivative of V with respect to capital X is equal to 1 minus 6 V minus V square whole upon 3 plus V. Now separating the variables V and X this implies 3 plus V whole upon 1 minus 6 V minus V square into d V is equal to dx over X. Now integrating both sides we get so integrating we have got this result which further implies integral 3 plus V whole upon V square plus 6 V minus 1 into d V is equal to minus integral dx over X. Now multiplying both sides by 2 this implies integral 2 V plus 6 whole upon V square plus 6 V minus 1 into d V is equal to minus 2 into integral dx over X. Now using this result, now here the derivative of V square plus 6 V minus 1 is 2 V plus 6 this means this is of the form integral of vectors of X over f of X dx which is equal to log of mod f of X. So this will be log of mod V square plus 6 V minus 1 which is equal to now here the integral of 1 minus X into dx is equal to log of mod X. So this will be minus 2 into log of mod X plus C where C is a constant. Now putting V is equal to capital Y over capital X in above equation we get log of mod capital Y square plus 6 into capital Y into capital X minus capital X square whole upon capital X square is equal to minus 2 into log of mod of capital X plus this is equal to capital X plus H which means capital X is equal to X minus H and here capital Y is equal to Y minus K where this is the value of H and this is the value of K. So this is the required solution of the given differential equation when capital X is equal to X minus H let us X minus 17 by 10 and capital Y is equal to Y minus K that is Y minus 19 by 10. So we have discussed an example for the case when A over capital A is not equal to V over capital B. Now let us discuss one exceptional case when A over capital A is equal to B over capital B is equal to a constant say R and this equation. Now let us discuss this case with the help of an example. Now comparing this equation with the equation that is with this differential equation where A over capital A will be 2 by 4 which is 1 by 2 is equal to B over capital B which is also 1 by 2. So let us solve this differential equation and let us see how to reduce it in homogeneous form. Now let us start with the solution here R let us A over capital A which is equal to B over capital B is equal to 1 by 2. Therefore the given equation can be written as dy over dx is equal to 1 by 2 into 4 X plus 4 Y the whole plus 2 whole upon 4 X plus 4 Y the whole plus 1. Now let it be equation A now putting 4 X plus 4 Y is equal to Z. So differentiating this equation with respect to X we are obtaining dy by dx is equal to 1 by 4 into dz by dx minus 4 the whole. So putting all these values in equation A we get now dy by dx is equal to 1 by 4 into dz by dx minus 4 the whole is equal to 1 by 2 into now 4 X plus 4 Y is equal to Z. So this will become 1 by 2 into Z plus 2 whole upon Z plus 1 the whole. Now on solving this equation this implies dz over dx is equal to 2 into Z plus 2 whole upon Z plus 1 the whole plus 4 which implies dz over dx is equal to 2 Z plus 4 plus 4 Z plus 4. Whole upon Z plus 1 which is equal to 6 Z plus 8 whole upon Z plus 1. Now separating the variables Z and X this implies Z plus 1 whole upon 6 Z plus 8 into dz is equal to dx. This further implies now taking 6 common from the denominator it will be 1 by 6 into 9 plus 1 whole upon Z plus 8 by 6 the whole into dz is equal to dx. Now here 2 into 4 is 8 and 2 into 3 is 6. Now integrating both sides we get 1 by 6 into integral Z plus 1 whole upon Z plus 4 by 3 into dz is equal to integral Tx. Now this implies integral now on dividing Z plus 1 by Z plus 4 by 3 that is here. This can be written as 1 minus 1 by 3 into 1 over Z plus 4 by 3 the whole into dz is equal to 6 into integral dx. This further implies Z minus 1 by 3 log of mod Z plus 4 by 3 is equal to 6X plus C. Now we have taken Z is equal to 4X plus 4 Y. So putting this value here this implies 4X plus 4 Y minus 1 by 3 into log of mod Z plus 4 by 3. Mod 4X plus 4 Y plus 4 by 3 is equal to 6X plus C and further on solving this implies 4 Y minus 2X the whole minus 1 by 3 into log of mod 4X plus 4 Y plus 4 by 3 is equal to C and this is the required solution of the given differential equation. So this is how we can solve this exception case. So in this session we have learnt about equations reducible to homogeneous equations and this completes our session. Hope you all have enjoyed the session.