Solve (d^2 y)/(dt^2 )-dy/dt-2y=0 given y(0)=-2 y^' (0)=5
Solve (d^2 y)/(dt^2 )-dy/dt-2y=0 given y(0)=-2 y^' (0)=5
TAM3B - Differential equations
( x - y ) dx -( x + y ) dy =0 Exact Equation If the first order partial derivatives of M( x,y ) and N( x,y ) are continuous then M dx + N dy = 0 is an exact equation if and only if ∂ M/ ∂ y = ∂ N/∂ x Hence to solve the exact equation M dx + N dy = 0 Integrate M with respect to x keeping y constant Integrate those terms in N not containing x with respect to y The sum of those two integrals equated to c is the solution. EXAMPLE 1 ( x^2-2xy+3y^2 )dx+(y^2+6xy-x^2 )dy=0 EXAMPLE 2 ( 2xy-sec^2x )dx+(x^2+2y)dy=0 EXAMPLE 3 Find the value of ‘ n ’ for which ( x+ye^2xy )dx+nxe^2xy dy=0 and solve it EXAMPLE 4 ( sin x. tan y+1)dx-(cosx.sec^2y )dy=0 EXAMPLE 5 ( e^y+1) cos x dx+e^y sinx dy=0
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